├── .gitconfig ├── README.md ├── files.proj ├── pictures ├── 3个非重复子数组最大和.png ├── 4S分析法 - 存储.png ├── 4S分析法 - 数据结构.png ├── 4S分析法.png ├── 位运算 ├── 位运算1.png ├── 位运算2.png ├── 位运算3.png ├── 位运算4.png ├── 位运算5.png ├── 位运算6.png ├── 位运算7.png ├── 位运算8.png ├── 原子计数.png ├── 字典里面的最长单词 - 题解.png ├── 字典里面的最长单词.png ├── 找二叉树最底层最左边的节点-样例.png ├── 找二叉树最底层最左边的节点.题解.png ├── 服务.png ├── 线段树1.png ├── 线段树2.png ├── 线段树3.png ├── 线段树4.png ├── 线段树5.png ├── 线段树6.png └── 线段树7.png ├── 令狐老师的算法小抄 └── 面试常考算法模板 Cheat Sheet V4.3.pdf ├── 其他 ├── n-gram-map-reduce.md └── readme.md ├── 动态规划 ├── arithmetic-slices-ii-subsequence.md ├── backpack-ii.md ├── backpack-iii.md ├── backpack-iv.md ├── backpack-v.md ├── backpack-vi.md ├── backpack.md ├── binary-tree-maximum-path-sum.md ├── bomb-enemy.md ├── burst-ballons.md ├── burst-balloons.md ├── calculation-the-sum-of-path.md ├── can-i-win.md ├── card-game.md ├── climbing-stairs-ii.md ├── climbing-stairs.md ├── coin-change.md ├── coins-in-a-line-ii.md ├── coins-in-a-line-iii.md ├── coins-in-a-line.md ├── copy-books-ii.md ├── copy-books.md ├── copy-list-with-random-pointer.md ├── count-numbers-with-unique-digits.md ├── count-square-submatrices-with-all-ones.md ├── create-maximum-number.md ├── decode-ways-ii.md ├── decode-ways.md ├── dices-sum.md ├── distinct-subsequences-ii.md ├── distinct-subsequences.md ├── drop-eggs-ii.md ├── drop-eggs.md ├── edit-distance.md ├── frog-jump.md ├── house-robber-ii.md ├── house-robber-iii.md ├── house-robber.md ├── interleaving-string.md ├── jump-game.md ├── k-edit-distance.md ├── largest-divisible-subset.md ├── longest-common-subsequence.md ├── longest-continuous-increasing-subsequence-ii.md ├── longest-continuous-increasing-subsequence.md ├── longest-increasing-continuous-subsequence.md ├── longest-increasing-subsequence.md ├── longest-palindromic-subsequence.md ├── longest-path-on-the-tree.md ├── longest-repeating-subsequence.md ├── longest-valid-parentheses.md ├── maximal-rectangle.md ├── maximal-square-ii.md ├── maximal-square.md ├── maximum-product-subarray.md ├── maximum-subarray-difference.md ├── maximum-subarray-ii.md ├── maximum-subarray-iii.md ├── maximum-subarray-iv.md ├── maximum-subarray-v.md ├── maximum-subarray.md ├── minimum-adjustment-cost.md ├── minimum-cost-to-merge-stones.md ├── minimum-falling-path-sum.md ├── minimum-path-sum.md ├── modern-ludo-i.md ├── non-negative-integers-without-consecutive-ones.md ├── palindrome-partitioning-ii.md ├── partition-equal-subset-sum.md ├── perfect-squares.md ├── post-office-problem.md ├── profitable-schemes.md ├── range-sum-query-2d-immutable.md ├── regular-expression-matching.md ├── russian-doll-envelopes.md ├── scramble-string.md ├── stone-game-ii.md ├── stone-game.md ├── traveling-salesman-problem.md ├── unique-paths-ii.md ├── unique-paths.md ├── unique-substrings-in-wraparound-string.md ├── word-break-ii.md ├── word-break-iii.md └── word-break.md ├── 图论 ├── find-the-connected-component-in-the-undirected-graph.md ├── 宽度优先搜索 │ ├── best-time-to-buy-and-sell-stock-ii.md │ ├── best-time-to-buy-and-sell-stock-iii.md │ ├── best-time-to-buy-and-sell-stock-iv.md │ ├── best-time-to-buy-and-sell-stock.md │ ├── binary-tree-zigzag-level-order-traversal.md │ ├── build-post-office-ii.md │ ├── convert-binary-tree-to-linked-lists-by-depth.md │ ├── knight-shortest-path-ii.md │ ├── knight-shortest-path.md │ ├── maximum-subarray-difference.md │ ├── path-sum.md │ ├── remove-substrings.md │ ├── route-between-two-nodes-in-graph.md │ ├── search-graph-nodes.md │ ├── shortest-path-to-the-destination.md │ ├── shortest-path-visiting-all-nodes.md │ ├── similar-string-groups.md │ ├── six-degrees.md │ ├── sliding-puzzle-ii.md │ ├── surrounded-regions.md │ ├── the-maze-iii.md │ ├── topological-sorting.md │ ├── tree.md │ ├── walls-and-gates.md │ ├── word-ladder-ii.md │ ├── word-ladder.md │ └── zombie-in-matrix.md └── 深度优先搜索 │ ├── add-operators.md │ ├── balanced-binary-tree.md │ ├── binary-tree-longest-consecutive-sequence-ii.md │ ├── binary-tree-longest-consecutive-sequence-iii.md │ ├── binary-tree-longest-consecutive-sequence.md │ ├── binary-tree-maximum-node.md │ ├── binary-tree-maximum-path-sum-ii.md │ ├── binary-tree-maximum-path-sum.md │ ├── binary-tree-path-sum-ii.md │ ├── binary-tree-path-sum-iii.md │ ├── binary-tree-path-sum.md │ ├── clone-binary-tree.md │ ├── clone-graph.md │ ├── combination-sum-ii.md │ ├── combination-sum-iv.md │ ├── combination-sum.md │ ├── combinations.md │ ├── continuous-subarray-sum-ii.md │ ├── continuous-subarray-sum.md │ ├── convert-sorted-array-to-binary-search-tree-with-minimal-height.md │ ├── convert-sorted-array-to-binary-search-tree.md │ ├── convert-sorted-list-to-balanced-bst.md │ ├── convert-sorted-list-to-binary-search-tree.md │ ├── decode-string.md │ ├── driving-problem.md │ ├── expression-evaluation.md │ ├── factorization.md │ ├── find-the-missing-number-ii.md │ ├── flatten-2d-vector.md │ ├── flatten-binary-tree-to-linked-list.md │ ├── flatten-list.md │ ├── flatten-nested-iterate.md │ ├── flatten-nested-list-iterator.md │ ├── flip-game-ii.md │ ├── increasing-subsequences.md │ ├── k-edit-distance.md │ ├── k-sum-ii.md │ ├── k-sum.md │ ├── keys-and-rooms.md │ ├── letter-combinations-of-a-phone-number-ii.md │ ├── letter-combinations-of-a-phone-number.md │ ├── longest-consecutive-sequence.md │ ├── maximum-depth-of-binary-tree.md │ ├── maximum-subarray-difference.md │ ├── maximum-subarray-ii.md │ ├── maximum-subarray-iii.md │ ├── maximum-subarray-iv.md │ ├── maximum-subarray-v.md │ ├── maximum-subarray.md │ ├── maximum-submatrix.md │ ├── maximum-subtree.md │ ├── minimum-depth-of-binary-tree.md │ ├── minimum-subarray.md │ ├── minimum-subtree.md │ ├── n-queens-ii.md │ ├── n-queens.md │ ├── nested-list-weight-sum.md │ ├── next-permutation-ii.md │ ├── next-permutation.md │ ├── number-of-islands.md │ ├── palindrome-partitioning.md │ ├── pascals-triangle.md │ ├── path-sum-ii.md │ ├── permutation-sequence.md │ ├── permutations-ii.md │ ├── permutations.md │ ├── populating-next-right-pointers-in-each-node-ii.md │ ├── prime-factorization.md │ ├── reconstruct-itinerary.md │ ├── recover-binary-search-tree.md │ ├── remove-node-in-binary-search-tree.md │ ├── string-permutation-ii.md │ ├── subsets-ii.md │ ├── subsets.md │ ├── subtree.md │ ├── sudoku-solver.md │ ├── sum-of-left-leaves.md │ ├── sum-root-to-leaf-numbers.md │ ├── symmetric-binary-tree.md │ ├── target-sum.md │ ├── triangle.md │ ├── webpage-crawler.md │ ├── wildcard-matching.md │ ├── word-break-ii.md │ ├── word-break.md │ ├── word-count.md │ ├── word-ladder-ii.md │ ├── word-ladder.md │ └── word-search-ii.md ├── 数据结构 ├── balanced-binary-tree.md ├── count-of-smaller-numbers-after-self.md ├── data-stream-median.md ├── find-median-from-data-stream.md ├── implement-queue-by-two-stacks.md ├── implement-stack-by-two-queues.md ├── kth-largest-element-ii.md ├── kth-largest-element.md ├── kth-largest-in-n-arrays.md ├── kth-smallest-number-in-sorted-matrix.md ├── kth-smallest-sum-in-two-sorted-arrays.md ├── largest-number.md ├── moving-average-from-data-stream.md ├── print-numbers-by-recursion.md ├── range-sum-query-2d-mutable.md ├── range-sum-query-mutable.md ├── segment-tree-build-ii.md ├── segment-tree-build.md ├── segment-tree-modify.md ├── segment-tree-query-ii.md ├── segment-tree-query.md ├── sequence-reconstruction.md ├── ⼆叉搜索树 │ ├── add-binary.md │ ├── balanced-binary-tree.md │ ├── binary-representation.md │ ├── binary-search-tree-iterator.md │ ├── binary-search.md │ ├── binary-tree-flipping.md │ ├── binary-tree-inorder-traversal.md │ ├── binary-tree-leaf-sum.md │ ├── binary-tree-leaves-order-traversal.md │ ├── binary-tree-level-order-traversal-ii.md │ ├── binary-tree-level-order-traversal.md │ ├── binary-tree-level-sum.md │ ├── binary-tree-longest-consecutive-sequence-ii.md │ ├── binary-tree-longest-consecutive-sequence-iii.md │ ├── binary-tree-longest-consecutive-sequence.md │ ├── binary-tree-maximum-node.md │ ├── binary-tree-maximum-path-sum-ii.md │ ├── binary-tree-maximum-path-sum.md │ ├── binary-tree-path-sum-ii.md │ ├── binary-tree-path-sum-iii.md │ ├── binary-tree-path-sum.md │ ├── binary-tree-paths.md │ ├── binary-tree-postorder-traversal.md │ ├── binary-tree-preorder-traversal.md │ ├── binary-tree-right-side-view.md │ ├── binary-tree-serialization.md │ ├── binary-tree-vertical-order-traversal.md │ ├── binary-tree-zigzag-level-order-traversal.md │ ├── binary-watch.md │ ├── boundary-of-binary-tree.md │ ├── classical-binary-search.md │ ├── clone-binary-tree.md │ ├── closest-binary-search-tree-value-ii.md │ ├── closest-binary-search-tree-value.md │ ├── complete-binary-tree.md │ ├── construct-binary-tree-from-inorder-and-postorder-traversal.md │ ├── construct-binary-tree-from-preorder-and-inorder-traversal.md │ ├── convert-binary-search-tree-to-doubly-linked-list.md │ ├── convert-binary-tree-to-linked-lists-by-depth.md │ ├── convert-bst-to-greater-tree.md │ ├── convert-sorted-array-to-binary-search-tree-with-minimal-height.md │ ├── convert-sorted-array-to-binary-search-tree.md │ ├── convert-sorted-list-to-binary-search-tree.md │ ├── count-1-in-binary.md │ ├── flatten-binary-tree-to-linked-list.md │ ├── identical-binary-tree.md │ ├── inorder-successor-in-binary-search-tree.md │ ├── inorder-successor-in-bst.md │ ├── insert-node-in-a-binary-search-tree.md │ ├── insert-node-in-binary-search-tree.md │ ├── invert-binary-tree.md │ ├── kth-smallest-element-in-a-bst.md │ ├── largest-bst-subtree.md │ ├── maximum-depth-of-binary-tree.md │ ├── minimum-depth-of-binary-tree.md │ ├── recover-binary-search-tree.md │ ├── remove-node-in-binary-search-tree.md │ ├── search-range-in-binary-search-tree.md │ ├── serialize-and-deserialize-binary-tree.md │ ├── symmetric-binary-tree.md │ ├── tweaked-identical-binary-tree.md │ ├── unique-binary-search-trees-ii.md │ ├── unique-binary-search-trees.md │ ├── validate-binary-search-tree.md │ └── verify-preorder-serialization-of-a-binary-tree.md ├── 二叉树分治 │ ├── balanced-binary-tree.md │ ├── binary-tree-flipping.md │ ├── binary-tree-inorder-traversal.md │ ├── binary-tree-leaf-sum.md │ ├── binary-tree-leaves-order-traversal.md │ ├── binary-tree-level-order-traversal-ii.md │ ├── binary-tree-level-order-traversal.md │ ├── binary-tree-level-sum.md │ ├── binary-tree-longest-consecutive-sequence-ii.md │ ├── binary-tree-longest-consecutive-sequence-iii.md │ ├── binary-tree-longest-consecutive-sequence.md │ ├── binary-tree-maximum-node.md │ ├── binary-tree-maximum-path-sum-ii.md │ ├── binary-tree-maximum-path-sum.md │ ├── binary-tree-path-sum-ii.md │ ├── binary-tree-path-sum-iii.md │ ├── binary-tree-path-sum.md │ ├── binary-tree-paths.md │ ├── binary-tree-postorder-traversal.md │ ├── binary-tree-preorder-traversal.md │ ├── binary-tree-right-side-view.md │ ├── binary-tree-serialization.md │ ├── binary-tree-vertical-order-traversal.md │ ├── binary-tree-zigzag-level-order-traversal.md │ ├── boundary-of-binary-tree.md │ ├── clone-binary-tree.md │ ├── complete-binary-tree.md │ ├── construct-binary-tree-from-inorder-and-postorder-traversal.md │ ├── construct-binary-tree-from-preorder-and-inorder-traversal.md │ ├── convert-binary-tree-to-linked-lists-by-depth.md │ ├── flatten-binary-tree-to-linked-list.md │ ├── identical-binary-tree.md │ ├── invert-binary-tree.md │ ├── maximum-depth-of-binary-tree.md │ ├── minimum-depth-of-binary-tree.md │ ├── serialize-and-deserialize-binary-tree.md │ ├── symmetric-binary-tree.md │ ├── tweaked-identical-binary-tree.md │ └── verify-preorder-serialization-of-a-binary-tree.md ├── 哈希表 │ ├── anagrams.md │ ├── binary-tree-vertical-order-traversal.md │ ├── binary-tree-zigzag-level-order-traversal.md │ ├── building-outline.md │ ├── clone-graph.md │ ├── consistent-hashing-ii.md │ ├── consistent-hashing.md │ ├── copy-list-with-random-pointer.md │ ├── count-characters.md │ ├── counting-bloom-filter.md │ ├── data-stream-as-disjoint-intervals.md │ ├── find-all-anagrams-in-a-string.md │ ├── first-unique-number-in-data-stream-ii.md │ ├── first-unique-number-in-data-stream.md │ ├── happy-number.md │ ├── hash-function.md │ ├── heart-beat.md │ ├── insert-delete-getrandom-o1-duplicates-allowed.md │ ├── insert-delete-getrandom-o1.md │ ├── island-perimeter.md │ ├── line-reflection.md │ ├── load-balancer.md │ ├── longest-palindrome.md │ ├── majority-number-ii.md │ ├── majority-number-iii.md │ ├── majority-number.md │ ├── max-points-on-a-line.md │ ├── memcache.md │ ├── mini-cassandra.md │ ├── minimum-window-substring.md │ ├── mirror-numbers.md │ ├── name-deduplication.md │ ├── number-of-boomerangs.md │ ├── rehashing.md │ ├── same-number.md │ ├── standard-bloom-filter.md │ ├── strstr-ii.md │ ├── subarray-sum-equals-k-ii.md │ ├── subarray-sum-equals-to-k-ii.md │ ├── tiny-url-ii.md │ ├── tiny-url.md │ ├── top-k-frequent-words.md │ ├── word-search-ii.md │ └── word-search.md ├── 堆 │ ├── heapify.md │ ├── high-five.md │ ├── k-closest-points.md │ ├── kth-largest-element-ii.md │ ├── kth-largest-element.md │ ├── kth-largest-in-n-arrays.md │ ├── kth-smallest-number-in-sorted-matrix.md │ ├── largest-number.md │ ├── merge-k-sorted-arrays.md │ ├── merge-k-sorted-interval-lists.md │ ├── merge-k-sorted-lists.md │ ├── shortest-subarray-with-sum-at-least-k.md │ ├── super-ugly-number.md │ ├── the-skyline-problem.md │ ├── top-k-largest-numbers.md │ ├── trapping-rain-water-ii.md │ └── ugly-number-ii.md ├── 字典树 │ ├── add-and-search-word-data-structure-design.md │ ├── add-and-search-word.md │ ├── boggle-game.md │ ├── implement-trie-prefix-tree.md │ ├── implement-trie.md │ └── word-squares.md ├── 栈 │ ├── 132-pattern.md │ ├── 132-patterns.md │ ├── basic-calculator-iii.md │ ├── basic-calculator.md │ ├── convert-expression-to-polish-notation.md │ ├── convert-expression-to-reverse-polish-notation.md │ ├── evaluate-reverse-polish-notation.md │ ├── expression-tree-build.md │ ├── implement-three-stacks-by-single-array.md │ ├── kth-smallest-sum-in-two-sorted-arrays.md │ ├── largest-rectangle-in-histogram.md │ ├── longest-valid-parentheses.md │ ├── max-stack.md │ ├── max-tree.md │ ├── meeting-rooms.md │ ├── min-stack.md │ ├── mock-hanoi-tower-by-stacks.md │ ├── odd-even-jump.md │ ├── set-of-stacks-ii.md │ ├── set-of-stacks.md │ ├── simplify-path.md │ ├── stack-sorting.md │ └── valid-parentheses.md └── 链表 │ ├── add-two-numbers-ii.md │ ├── add-two-numbers.md │ ├── animal-shelter.md │ ├── convert-array-list-to-linked-list.md │ ├── convert-binary-search-tree-to-doubly-linked-list.md │ ├── convert-binary-tree-to-linked-lists-by-depth.md │ ├── convert-linked-list-to-array-list.md │ ├── count-linked-list-nodes.md │ ├── delete-node-in-the-middle-of-singly-linked-list.md │ ├── find-node-in-linked-list.md │ ├── implement-queue-by-linked-list-ii.md │ ├── implement-queue-by-linked-list.md │ ├── insert-into-a-cyclic-sorted-list.md │ ├── insert-node-in-sorted-linked-list.md │ ├── insertion-sort-list.md │ ├── intersection-of-two-linked-lists.md │ ├── merge-k-sorted-lists.md │ ├── merge-two-sorted-lists.md │ ├── middle-of-linked-list.md │ ├── nth-to-last-node-in-list.md │ ├── palindrome-linked-list.md │ ├── remove-duplicates-from-sorted-list-ii.md │ ├── remove-duplicates-from-sorted-list.md │ ├── remove-duplicates-from-unsorted-list.md │ ├── remove-linked-list-elements.md │ ├── remove-nth-node-from-end-of-list.md │ ├── reorder-list.md │ ├── reverse-linked-list-ii.md │ ├── reverse-linked-list.md │ ├── reverse-nodes-in-k-group.md │ ├── rotate-list.md │ ├── sort-list.md │ ├── swap-nodes-in-pairs.md │ └── swap-two-nodes-in-linked-list.md ├── 算法与数据结构 ├── 位运算面试题总结.md ├── 动态规划十问十答.md └── 线段树知识点总结.md ├── 算法思维题 ├── lowest-common-ancestor-ii.md ├── lowest-common-ancestor-iii.md ├── lowest-common-ancestor.md ├── permutation-sequence.md ├── 二分法 │ ├── build-post-office.md │ ├── classical-binary-search.md │ ├── closest-number-in-sorted-array.md │ ├── continuous-subarray-sum-ii.md │ ├── count-of-smaller-number-before-itself.md │ ├── count-of-smaller-number.md │ ├── divide-two-integers.md │ ├── dungeon-game.md │ ├── find-k-closest-elements.md │ ├── find-minimum-in-rotated-sorted-array-ii.md │ ├── find-minimum-in-rotated-sorted-array.md │ ├── find-peak-element-ii.md │ ├── find-peak-element.md │ ├── find-right-interval.md │ ├── first-bad-version.md │ ├── first-position-of-target.md │ ├── geohash-ii.md │ ├── geohash.md │ ├── h-index-ii.md │ ├── h-index.md │ ├── intersection-of-two-arrays-ii.md │ ├── intersection-of-two-arrays.md │ ├── interval-minimum-number.md │ ├── interval-sum-ii.md │ ├── interval-sum.md │ ├── k-closest-numbers-in-sorted-array.md │ ├── last-position-of-target.md │ ├── maximum-absolute-value.md │ ├── maximum-average-subarray-ii.md │ ├── maximum-average-subarray.md │ ├── maximum-number-in-mountain-sequence.md │ ├── median-of-k-sorted-arrays.md │ ├── powx-n.md │ ├── search-a-2d-matrix.md │ ├── search-for-a-range.md │ ├── search-in-a-big-sorted-array.md │ ├── search-insert-position.md │ ├── smallest-rectangle-enclosing-black-pixels.md │ ├── split-array-largest-sum.md │ ├── sqrt-x.md │ ├── sqrtx-ii.md │ ├── subarray-sum-ii.md │ ├── total-occurrence-of-target.md │ ├── web-logger.md │ ├── wiggle-sort-ii.md │ └── wood-cut.md ├── 位运算 │ ├── a-b-problem.md │ ├── add-binary.md │ ├── add-digits.md │ ├── add-strings.md │ ├── binary-representation.md │ ├── binary-watch.md │ ├── bitwise-and-of-numbers-range.md │ ├── convert-a-number-to-hexadecimal.md │ ├── count-1-in-binary.md │ ├── flip-bits.md │ ├── gray-code.md │ ├── hamming-distance.md │ ├── number-of-1-bits.md │ ├── o1-check-power-of-2.md │ ├── single-number-ii.md │ ├── single-number-iii.md │ ├── single-number.md │ ├── swap-bits.md │ ├── swap-without-extra-variable-only-c.md │ └── update-bits.md ├── 其他 │ ├── alien-dictionary.md │ ├── amicable-pair.md │ ├── assignment-operator-overloading-c-only.md │ ├── convex-polygon.md │ ├── cosine-similarity.md │ ├── delete-and-earn.md │ ├── digit-counts.md │ ├── fibonacci.md │ ├── find-the-celebrity.md │ ├── find-the-missing-number.md │ ├── fizz-buzz.md │ ├── friendship-service.md │ ├── generate-parentheses.md │ ├── gfs-client.md │ ├── guess-number-game.md │ ├── insert-interval.md │ ├── kth-prime-number.md │ ├── lfu-cache.md │ ├── matrix-zigzag-traversal.md │ ├── max-of-3-numbers.md │ ├── narcissistic-number.md │ ├── next-smaller-and-larger-number-with-the-same-1-bits.md │ ├── previous-permutation.md │ ├── rectangle-overlap.md │ ├── remove-element.md │ ├── restore-ip-addresses.md │ ├── reverse-3-digit-integer.md │ ├── reverse-bits.md │ ├── reverse-integer.md │ ├── root-of-equation.md │ ├── rotate-array.md │ ├── search-in-rotated-sorted-array-ii.md │ ├── setter-and-getter.md │ ├── shape-factory.md │ ├── simple-calculator.md │ ├── singleton.md │ ├── strings-homomorphism.md │ ├── strings-serialization.md │ ├── student-id.md │ ├── student-level.md │ ├── substring-rotation.md │ ├── substring-with-concatenation-of-all-words.md │ ├── sudoku-solver.md │ ├── text-justification.md │ ├── toy-factory.md │ ├── trailing-zeros.md │ ├── trie-service.md │ ├── tweaked-identical-binary-tree.md │ ├── two-strings-are-anagrams.md │ ├── ugly-number.md │ ├── unique-binary-search-trees-ii.md │ ├── unique-binary-search-trees.md │ ├── utf-8-validation.md │ ├── valid-number.md │ ├── valid-sudoku.md │ ├── verify-preorder-serialization-of-a-binary-tree.md │ ├── word-count.md │ └── zigzag-conversion.md ├── 分治法 │ ├── fast-power.md │ ├── kth-smallest-numbers-in-unsorted-array.md │ ├── nuts-bolts-problem.md │ ├── sort-colors-ii.md │ ├── subtree-with-maximum-average.md │ ├── symmetric-tree.md │ └── validate-binary-search-tree.md ├── 双指针 │ ├── 3sum-closest.md │ ├── 3sum.md │ ├── 4sum-ii.md │ ├── 4sum.md │ ├── container-with-most-water.md │ ├── continuous-subarray-sum-ii.md │ ├── find-the-duplicate-number.md │ ├── interleaving-positive-and-negative-numbers.md │ ├── linked-list-cycle-ii.md │ ├── linked-list-cycle.md │ ├── longest-substring-with-at-most-k-distinct-characters.md │ ├── longest-substring-without-repeating-characters.md │ ├── maximum-submatrix.md │ ├── merge-sorted-array.md │ ├── minimum-size-subarray-sum.md │ ├── minimum-window-substring.md │ ├── move-zeroes.md │ ├── partition-array-by-odd-and-even.md │ ├── partition-list.md │ ├── remove-duplicate-numbers-in-array.md │ ├── sliding-window-maximum.md │ ├── sliding-window-median.md │ ├── sort-colors.md │ ├── subarray-sum-ii.md │ ├── the-smallest-difference.md │ ├── triangle-count.md │ ├── two-sum-closest-to-target.md │ ├── two-sum-closest.md │ ├── two-sum-difference-equals-to-target.md │ ├── two-sum-greater-than-target.md │ ├── two-sum-input-array-is-sorted.md │ ├── two-sum-less-than-or-equal-to-target.md │ ├── two-sum-unique-pairs.md │ ├── valid-palindrome-ii.md │ ├── valid-palindrome.md │ ├── valid-word-abbreviation.md │ ├── window-sum.md │ ├── zigzag-iterator-ii.md │ └── zigzag-iterator.md ├── 字典MAP │ ├── inverted-index-map-reduce.md │ ├── inverted-index.md │ ├── permutation-index-ii.md │ ├── string-permutation.md │ ├── substring-with-at-least-k-distinct-characters.md │ ├── top-k-frequent-words-ii.md │ ├── two-sum-iii-data-structure-design.md │ ├── two-sum.md │ ├── typeahead.md │ ├── unique-characters.md │ ├── word-abbreviation-set.md │ └── words-abbreviation.md ├── 并查集 │ ├── accounts-merge.md │ ├── connected-component-in-undirected-graph.md │ ├── connecting-graph-ii.md │ ├── connecting-graph-iii.md │ ├── connecting-graph.md │ ├── find-the-weak-connected-component-in-the-directed-graph.md │ ├── graph-valid-tree.md │ ├── minimum-spanning-tree.md │ ├── number-of-islands-ii.md │ └── number-of-islands.md ├── 排序算法 │ ├── data-stream-median.md │ ├── find-median-from-data-stream.md │ ├── insert-into-a-cyclic-sorted-list.md │ ├── insertion-sort-list.md │ ├── median-in-data-stream.md │ ├── median-of-k-sorted-arrays.md │ ├── median-of-two-sorted-arrays.md │ ├── median.md │ ├── meeting-rooms.md │ ├── merge-k-sorted-lists.md │ ├── merge-two-sorted-arrays.md │ ├── merge-two-sorted-interval-lists.md │ ├── merge-two-sorted-lists.md │ ├── nuts-bolts-problem.md │ ├── partition-array-by-odd-and-even.md │ ├── partition-array-ii.md │ ├── partition-array.md │ ├── recover-rotated-sorted-array.md │ ├── remove-duplicates-from-sorted-list.md │ ├── reverse-pairs.md │ ├── sliding-window-median.md │ ├── sort-integers-ii.md │ ├── sort-integers-map-reduce.md │ ├── sort-integers.md │ ├── sort-letters-by-case.md │ ├── sort-list.md │ ├── top-k-frequent-elements.md │ ├── top-k-frequent-words-map-reduce.md │ ├── top-k-largest-numbers.md │ └── wiggle-sort.md └── 贪心算法 │ ├── assign-cookies.md │ ├── course-schedule-ii.md │ ├── course-schedule-iii.md │ ├── course-schedule.md │ ├── delete-digits.md │ ├── gas-station.md │ ├── increasing-triplet-subsequence.md │ ├── jump-game-ii.md │ ├── maximum-gap.md │ ├── minimum-number-of-arrows-to-burst-balloons.md │ ├── queue-reconstruction-by-height.md │ └── russian-doll-envelopes.md ├── 算法面试题解 ├── Google 面试题 | 3个非重复子数组最大和.md ├── Google 面试题 | 原子计数.md ├── Google 面试题 | 合法的IP的地址.md ├── Google 面试题 | 字典里面的最长单词.md └── Google 面试题 | 找二叉树最底层最左边的节点.md ├── 系统设计 Syestem Design └── 如何答好面试中的系统设计题.md ├── 编程新手必刷50题 V1.0.docx ├── 面试常考算法必刷题库.pdf ├── 题解 ├── 132-pattern.md ├── 3sum-closest.md ├── 3sum.md ├── 4sum-ii.md ├── 4sum.md ├── a-b-problem.md ├── accounts-merge.md ├── add-and-search-word-data-structure-design.md ├── add-and-search-word.md ├── add-binary.md ├── add-digits.md ├── add-operators.md ├── add-strings.md ├── add-two-numbers-ii.md ├── add-two-numbers.md ├── alien-dictionary.md ├── amicable-pair.md ├── anagram-map-reduce.md ├── anagrams.md ├── animal-shelter.md ├── arithmetic-slices-ii-subsequence.md ├── arraylist.md ├── assign-cookies.md ├── assignment-operator-overloading-c-only.md ├── backpack-ii.md ├── backpack-iii.md ├── backpack-iv.md ├── backpack-v.md ├── backpack-vi.md ├── backpack.md ├── balanced-binary-tree.md ├── basic-calculator-iii.md ├── basic-calculator.md ├── battleships-in-a-board.md ├── best-time-to-buy-and-sell-stock-ii.md ├── best-time-to-buy-and-sell-stock-iii.md ├── best-time-to-buy-and-sell-stock-iv.md ├── best-time-to-buy-and-sell-stock.md ├── binary-representation.md ├── binary-search-tree-iterator.md ├── binary-tree-flipping.md ├── binary-tree-inorder-traversal.md ├── binary-tree-leaf-sum.md ├── binary-tree-leaves-order-traversal.md ├── binary-tree-level-order-traversal-ii.md ├── binary-tree-level-order-traversal.md ├── binary-tree-level-sum.md ├── binary-tree-longest-consecutive-sequence-ii.md ├── binary-tree-longest-consecutive-sequence-iii.md ├── binary-tree-longest-consecutive-sequence.md ├── binary-tree-maximum-node.md ├── binary-tree-maximum-path-sum-ii.md ├── binary-tree-maximum-path-sum.md ├── binary-tree-path-sum-ii.md ├── binary-tree-path-sum-iii.md ├── binary-tree-path-sum.md ├── binary-tree-paths.md ├── binary-tree-postorder-traversal.md ├── binary-tree-preorder-traversal.md ├── binary-tree-right-side-view.md ├── binary-tree-serialization.md ├── binary-tree-vertical-order-traversal.md ├── binary-tree-zigzag-level-order-traversal.md ├── binary-watch.md ├── bitwise-and-of-numbers-range.md ├── boggle-game.md ├── bomb-enemy.md ├── boundary-of-binary-tree.md ├── build-post-office-ii.md ├── build-post-office.md ├── building-outline.md ├── burst-ballons.md ├── burst-balloons.md ├── calculation-the-sum-of-path.md ├── can-i-win.md ├── can-place-flowers.md ├── candy.md ├── card-game.md ├── char-to-integer.md ├── classical-binary-search.md ├── climbing-stairs-ii.md ├── climbing-stairs.md ├── clone-binary-tree.md ├── clone-graph.md ├── closest-binary-search-tree-value-ii.md ├── closest-binary-search-tree-value.md ├── closest-number-in-sorted-array.md ├── coin-change.md ├── coins-in-a-line-ii.md ├── coins-in-a-line-iii.md ├── coins-in-a-line.md ├── combination-sum-ii.md ├── combination-sum-iv.md ├── combination-sum.md ├── combinations.md ├── compare-strings.md ├── complete-binary-tree.md ├── connected-component-in-undirected-graph.md ├── connecting-graph-ii.md ├── connecting-graph-iii.md ├── connecting-graph.md ├── consistent-hashing-ii.md ├── consistent-hashing.md ├── construct-binary-tree-from-inorder-and-postorder-traversal.md ├── construct-binary-tree-from-preorder-and-inorder-traversal.md ├── container-with-most-water.md ├── continuous-subarray-sum-ii.md ├── continuous-subarray-sum.md ├── convert-a-number-to-hexadecimal.md ├── convert-array-list-to-linked-list.md ├── convert-binary-search-tree-to-doubly-linked-list.md ├── convert-binary-tree-to-linked-lists-by-depth.md ├── convert-bst-to-greater-tree.md ├── convert-expression-to-polish-notation.md ├── convert-expression-to-reverse-polish-notation.md ├── convert-linked-list-to-array-list.md ├── convert-sorted-array-to-binary-search-tree-with-minimal-height.md ├── convert-sorted-array-to-binary-search-tree.md ├── convert-sorted-list-to-binary-search-tree.md ├── convex-polygon.md ├── copy-books-ii.md ├── copy-books.md ├── copy-list-with-random-pointer.md ├── cosine-similarity.md ├── count-1-in-binary.md ├── count-and-say.md ├── count-characters.md ├── count-linked-list-nodes.md ├── count-numbers-with-unique-digits.md ├── count-of-smaller-number-before-itself.md ├── count-of-smaller-number.md ├── count-of-smaller-numbers-after-self.md ├── count-square-submatrices-with-all-ones.md ├── counting-bloom-filter.md ├── course-schedule-ii.md ├── course-schedule-iii.md ├── course-schedule.md ├── create-maximum-number.md ├── data-stream-as-disjoint-intervals.md ├── data-stream-median.md ├── decode-string.md ├── decode-ways-ii.md ├── decode-ways.md ├── delete-and-earn.md ├── delete-digits.md ├── delete-node-in-the-middle-of-singly-linked-list.md ├── dices-sum.md ├── digit-counts.md ├── distinct-subsequences-ii.md ├── distinct-subsequences.md ├── divide-two-integers.md ├── driving-problem.md ├── drop-eggs-ii.md ├── drop-eggs.md ├── dungeon-game.md ├── edit-distance-ii.md ├── edit-distance.md ├── evaluate-reverse-polish-notation.md ├── excel-sheet-column-title.md ├── expression-evaluation.md ├── expression-expand.md ├── expression-tree-build.md ├── factorization.md ├── fast-power.md ├── fibonacci.md ├── find-all-anagrams-in-a-string.md ├── find-k-closest-elements.md ├── find-median-from-data-stream.md ├── find-minimum-in-rotated-sorted-array-ii.md ├── find-minimum-in-rotated-sorted-array.md ├── find-node-in-linked-list.md ├── find-peak-element-ii.md ├── find-peak-element.md ├── find-right-interval.md ├── find-the-celebrity.md ├── find-the-connected-component-in-the-undirected-graph.md ├── find-the-duplicate-number.md ├── find-the-missing-number-ii.md ├── find-the-missing-number.md ├── find-the-weak-connected-component-in-the-directed-graph.md ├── first-bad-version.md ├── first-missing-positive.md ├── first-position-of-target.md ├── first-position-unique-character.md ├── first-unique-character-in-a-string.md ├── first-unique-number-in-data-stream-ii.md ├── first-unique-number-in-data-stream.md ├── fizz-buzz.md ├── flatten-2d-vector.md ├── flatten-binary-tree-to-linked-list.md ├── flatten-list.md ├── flatten-nested-list-iterator.md ├── flip-bits.md ├── flip-game-ii.md ├── flip-string-to-monotone-increasing.md ├── friendship-service.md ├── frog-jump.md ├── gas-station.md ├── generate-arraylist-with-given-size.md ├── generate-parentheses.md ├── geohash-ii.md ├── geohash.md ├── gfs-client.md ├── graph-valid-tree.md ├── gray-code.md ├── guess-number-game.md ├── h-index-ii.md ├── h-index.md ├── hamming-distance.md ├── happy-number.md ├── hash-function.md ├── heapify.md ├── heart-beat.md ├── high-five.md ├── house-robber-ii.md ├── house-robber-iii.md ├── house-robber.md ├── identical-binary-tree.md ├── implement-deque-by-circulant-array.md ├── implement-queue-by-interface.md ├── implement-queue-by-linked-list-ii.md ├── implement-queue-by-linked-list.md ├── implement-queue-by-two-stacks.md ├── implement-stack-by-two-queues.md ├── implement-stack.md ├── implement-strstr.md ├── implement-three-stacks-by-single-array.md ├── implement-trie-prefix-tree.md ├── implement-trie.md ├── increasing-subsequences.md ├── increasing-triplet-subsequence.md ├── inorder-successor-in-binary-search-tree.md ├── inorder-successor-in-bst.md ├── insert-delete-getrandom-o1-duplicates-allowed.md ├── insert-delete-getrandom-o1.md ├── insert-interval.md ├── insert-into-a-cyclic-sorted-list.md ├── insert-node-in-a-binary-search-tree.md ├── insert-node-in-sorted-linked-list.md ├── insertion-sort-list.md ├── integer-to-english-words.md ├── integer-to-roman.md ├── interleaving-positive-and-negative-numbers.md ├── interleaving-string.md ├── intersection-of-arrays.md ├── intersection-of-two-arrays-ii.md ├── intersection-of-two-arrays.md ├── intersection-of-two-linked-lists.md ├── interval-minimum-number.md ├── interval-sum-ii.md ├── interval-sum.md ├── invert-binary-tree.md ├── inverted-index-map-reduce.md ├── inverted-index.md ├── is-subsequence.md ├── island-perimeter.md ├── jump-game-ii.md ├── jump-game.md ├── k-closest-numbers-in-sorted-array.md ├── k-closest-points.md ├── k-edit-distance.md ├── k-sum-ii.md ├── k-sum.md ├── keys-and-rooms.md ├── knight-shortest-path-ii.md ├── knight-shortest-path.md ├── kth-largest-element-ii.md ├── kth-largest-element.md ├── kth-largest-in-n-arrays.md ├── kth-prime-number.md ├── kth-smallest-element-in-a-bst.md ├── kth-smallest-number-in-sorted-matrix.md ├── kth-smallest-numbers-in-unsorted-array.md ├── kth-smallest-sum-in-two-sorted-arrays.md ├── largest-bst-subtree.md ├── largest-divisible-subset.md ├── largest-number.md ├── largest-rectangle-in-histogram.md ├── last-position-of-target.md ├── left-pad.md ├── length-of-last-word.md ├── letter-combinations-of-a-phone-number-ii.md ├── letter-combinations-of-a-phone-number.md ├── lfu-cache.md ├── line-reflection.md ├── linked-list-cycle-ii.md ├── linked-list-cycle.md ├── load-balancer.md ├── longest-common-prefix.md ├── longest-common-subsequence.md ├── longest-common-substring.md ├── longest-consecutive-sequence.md ├── longest-continuous-increasing-subsequence-ii.md ├── longest-continuous-increasing-subsequence.md ├── longest-increasing-continuous-subsequence.md ├── longest-increasing-subsequence.md ├── longest-palindrome.md ├── longest-palindromic-subsequence.md ├── longest-palindromic-substring.md ├── longest-path-on-the-tree.md ├── longest-repeating-subsequence.md ├── longest-substring-with-at-least-k-repeating-characters.md ├── longest-substring-with-at-most-k-distinct-characters.md ├── longest-substring-without-repeating-characters.md ├── longest-valid-parentheses.md ├── longest-words.md ├── lowercase-to-uppercase-ii.md ├── lowercase-to-uppercase.md ├── lowest-common-ancestor-ii.md ├── lowest-common-ancestor-iii.md ├── lowest-common-ancestor.md ├── lru-cache.md ├── majority-number-ii.md ├── majority-number-iii.md ├── majority-number.md ├── matrix-zigzag-traversal.md ├── max-of-3-numbers.md ├── max-of-array.md ├── max-points-on-a-line.md ├── max-stack.md ├── max-tree.md ├── maximal-rectangle.md ├── maximal-square-ii.md ├── maximal-square.md ├── maximum-absolute-value.md ├── maximum-average-subarray-ii.md ├── maximum-average-subarray.md ├── maximum-depth-of-binary-tree.md ├── maximum-gap.md ├── maximum-number-in-mountain-sequence.md ├── maximum-product-subarray.md ├── maximum-subarray-difference.md ├── maximum-subarray-ii.md ├── maximum-subarray-iii.md ├── maximum-subarray-iv.md ├── maximum-subarray-v.md ├── maximum-subarray.md ├── maximum-submatrix.md ├── maximum-subtree.md ├── median-of-k-sorted-arrays.md ├── median-of-two-sorted-arrays.md ├── median.md ├── meeting-rooms-ii.md ├── meeting-rooms.md ├── memcache.md ├── merge-intervals.md ├── merge-k-sorted-arrays.md ├── merge-k-sorted-interval-lists.md ├── merge-k-sorted-lists.md ├── merge-sorted-array-ii.md ├── merge-sorted-array.md ├── merge-two-sorted-arrays.md ├── merge-two-sorted-interval-lists.md ├── merge-two-sorted-lists.md ├── middle-of-linked-list.md ├── min-stack.md ├── mini-cassandra.md ├── mini-twitter.md ├── mini-uber.md ├── mini-yelp.md ├── minimum-adjustment-cost.md ├── minimum-cost-to-merge-stones.md ├── minimum-depth-of-binary-tree.md ├── minimum-falling-path-sum.md ├── minimum-moves-to-equal-array-elements-ii.md ├── minimum-moves-to-equal-array-elements.md ├── minimum-number-of-arrows-to-burst-balloons.md ├── minimum-path-sum.md ├── minimum-size-subarray-sum.md ├── minimum-spanning-tree.md ├── minimum-subarray.md ├── minimum-subtree.md ├── minimum-window-substring.md ├── mirror-numbers.md ├── missing-integer.md ├── missing-interval.md ├── mock-hanoi-tower-by-stacks.md ├── modern-ludo-i.md ├── monochrome-screen.md ├── move-zeroes.md ├── moving-average-from-data-stream.md ├── multiply-strings.md ├── n-gram-map-reduce.md ├── n-queens-ii.md ├── n-queens.md ├── name-deduplication.md ├── narcissistic-number.md ├── nested-list-weight-sum.md ├── next-permutation-ii.md ├── next-permutation.md ├── next-smaller-and-larger-number-with-the-same-1-bits.md ├── non-negative-integers-without-consecutive-ones.md ├── nth-to-last-node-in-list.md ├── null-terminated-string.md ├── number-of-1-bits.md ├── number-of-airplanes-in-the-sky.md ├── number-of-boomerangs.md ├── number-of-islands-ii.md ├── number-of-islands.md ├── number-of-ways-to-represent-n-cents.md ├── number-of-ways-to-stay-in-the-same-place-after-some-steps-ii.md ├── nuts-bolts-problem.md ├── o1-check-power-of-2.md ├── odd-even-jump.md ├── ones-and-zeroes.md ├── optimal-account-balancing.md ├── paint-fence.md ├── paint-house-ii.md ├── paint-house.md ├── palindrome-linked-list.md ├── palindrome-number.md ├── palindrome-partitioning-ii.md ├── palindrome-partitioning.md ├── parking-lot.md ├── partition-array-by-odd-and-even.md ├── partition-array-ii.md ├── partition-array.md ├── partition-equal-subset-sum.md ├── partition-list.md ├── pascals-triangle-ii.md ├── pascals-triangle.md ├── pass-interview.md ├── path-sum-ii.md ├── path-sum.md ├── perfect-squares.md ├── permutation-index-ii.md ├── permutation-index.md ├── permutation-sequence.md ├── permutations-ii.md ├── permutations.md ├── plus-one.md ├── populating-next-right-pointers-in-each-node-ii.md ├── post-office-problem.md ├── powx-n.md ├── previous-permutation.md ├── prime-factorization.md ├── print-numbers-by-recursion.md ├── product-of-array-exclude-itself.md ├── profitable-schemes.md ├── queue-reconstruction-by-height.md ├── range-sum-query-2d-immutable.md ├── range-sum-query-2d-mutable.md ├── range-sum-query-mutable.md ├── rat-jump.md ├── read-n-characters-given-read4-ii-call-multiple-times.md ├── reconstruct-itinerary.md ├── recover-binary-search-tree.md ├── recover-rotated-sorted-array.md ├── rectangle-area.md ├── rectangle-overlap.md ├── reference.md ├── regular-expression-matching.md ├── rehashing.md ├── remove-duplicate-numbers-in-array.md ├── remove-duplicates-from-sorted-array-ii.md ├── remove-duplicates-from-sorted-array.md ├── remove-duplicates-from-sorted-list-ii.md ├── remove-duplicates-from-sorted-list.md ├── remove-duplicates-from-unsorted-list.md ├── remove-element.md ├── remove-linked-list-elements.md ├── remove-node-in-binary-search-tree.md ├── remove-nth-node-from-end-of-list.md ├── remove-substrings.md ├── reorder-array-to-construct-the-minimum-number.md ├── reorder-list.md ├── repeated-dna-sequences.md ├── repeated-substring-pattern.md ├── restore-ip-addresses.md ├── reverse-3-digit-integer.md ├── reverse-bits.md ├── reverse-integer.md ├── reverse-linked-list-ii.md ├── reverse-linked-list.md ├── reverse-nodes-in-k-group.md ├── reverse-pairs.md ├── reverse-words-in-a-string.md ├── roman-to-integer.md ├── root-of-equation.md ├── rotate-array.md ├── rotate-image.md ├── rotate-list.md ├── rotate-string.md ├── route-between-two-nodes-in-graph.md ├── russian-doll-envelopes.md ├── same-number.md ├── scramble-string.md ├── search-a-2d-matrix-ii.md ├── search-a-2d-matrix.md ├── search-for-a-range.md ├── search-graph-nodes.md ├── search-in-a-big-sorted-array.md ├── search-in-rotated-sorted-array-ii.md ├── search-in-rotated-sorted-array.md ├── search-insert-position.md ├── search-range-in-binary-search-tree.md ├── second-max-of-array.md ├── segment-tree-build-ii.md ├── segment-tree-build.md ├── segment-tree-modify.md ├── segment-tree-query-ii.md ├── segment-tree-query.md ├── sequence-reconstruction.md ├── serialize-and-deserialize-binary-tree.md ├── set-matrix-zeroes.md ├── set-of-stacks-ii.md ├── set-of-stacks.md ├── setter-and-getter.md ├── shape-factory.md ├── shortest-palindrome.md ├── shortest-path-to-the-destination.md ├── shortest-path-visiting-all-nodes.md ├── shortest-subarray-with-sum-at-least-k.md ├── similar-string-groups.md ├── simple-calculator.md ├── simplify-path.md ├── single-number-ii.md ├── single-number-iii.md ├── single-number.md ├── singleton.md ├── six-degrees.md ├── sliding-puzzle-ii.md ├── sliding-window-matrix-maximum.md ├── sliding-window-maximum.md ├── sliding-window-median.md ├── smallest-rectangle-enclosing-black-pixels.md ├── sort-colors-ii.md ├── sort-colors.md ├── sort-integers-ii.md ├── sort-integers-map-reduce.md ├── sort-integers.md ├── sort-letters-by-case.md ├── sort-list.md ├── space-replacement.md ├── sparse-matrix-multiplication.md ├── spiral-matrix-ii.md ├── spiral-matrix.md ├── split-array-largest-sum.md ├── sqrt-x.md ├── sqrtx-ii.md ├── stack-sorting.md ├── standard-bloom-filter.md ├── stone-game-ii.md ├── stone-game.md ├── string-compression.md ├── string-permutation-ii.md ├── string-permutation.md ├── string-to-integer-atoi.md ├── string-to-integer.md ├── strings-homomorphism.md ├── strings-serialization.md ├── strstr-ii.md ├── student-id.md ├── student-level.md ├── subarray-sum-closest.md ├── subarray-sum-equals-k-ii.md ├── subarray-sum-equals-to-k-ii.md ├── subarray-sum-ii.md ├── subarray-sum.md ├── submatrix-sum.md ├── subsets-ii.md ├── subsets.md ├── substring-rotation.md ├── substring-with-at-least-k-distinct-characters.md ├── substring-with-concatenation-of-all-words.md ├── subtree-with-maximum-average.md ├── subtree.md ├── sudoku-solver.md ├── sum-of-left-leaves.md ├── sum-root-to-leaf-numbers.md ├── super-ugly-number.md ├── surrounded-regions.md ├── swap-bits.md ├── swap-nodes-in-pairs.md ├── swap-two-integers-in-array.md ├── swap-two-nodes-in-linked-list.md ├── swap-without-extra-variable-only-c.md ├── symmetric-binary-tree.md ├── symmetric-tree.md ├── system-longest-file-path.md ├── target-sum.md ├── text-justification.md ├── the-maze-iii.md ├── the-skyline-problem.md ├── the-smallest-difference.md ├── tiny-url-ii.md ├── tiny-url.md ├── top-k-frequent-elements.md ├── top-k-frequent-words-ii.md ├── top-k-frequent-words-map-reduce.md ├── top-k-frequent-words.md ├── top-k-largest-numbers.md ├── topological-sorting.md ├── total-occurrence-of-target.md ├── toy-factory.md ├── trailing-zeros.md ├── trapping-rain-water-ii.md ├── trapping-rain-water.md ├── traveling-salesman-problem.md ├── tree.md ├── triangle-count.md ├── triangle.md ├── trie-service.md ├── tweaked-identical-binary-tree.md ├── two-strings-are-anagrams.md ├── two-sum-closest-to-target.md ├── two-sum-closest.md ├── two-sum-difference-equals-to-target.md ├── two-sum-greater-than-target.md ├── two-sum-iii-data-structure-design.md ├── two-sum-input-array-is-sorted.md ├── two-sum-less-than-or-equal-to-target.md ├── two-sum-unique-pairs.md ├── two-sum.md ├── typeahead.md ├── ugly-number-ii.md ├── ugly-number.md ├── unique-binary-search-trees-ii.md ├── unique-binary-search-trees.md ├── unique-characters.md ├── unique-paths-ii.md ├── unique-paths.md ├── unique-substrings-in-wraparound-string.md ├── update-bits.md ├── url-parser.md ├── utf-8-validation.md ├── valid-number.md ├── valid-palindrome-ii.md ├── valid-palindrome.md ├── valid-parentheses.md ├── valid-sudoku.md ├── valid-word-abbreviation.md ├── validate-binary-search-tree.md ├── validate-ip-address.md ├── verify-preorder-serialization-of-a-binary-tree.md ├── walls-and-gates.md ├── web-logger.md ├── webpage-crawler.md ├── wiggle-sort-ii.md ├── wiggle-sort.md ├── wildcard-matching.md ├── window-sum.md ├── wood-cut.md ├── word-abbreviation-set.md ├── word-break-ii.md ├── word-break-iii.md ├── word-break.md ├── word-count.md ├── word-ladder-ii.md ├── word-ladder.md ├── word-pattern-ii.md ├── word-search-ii.md ├── word-search.md ├── word-squares.md ├── words-abbreviation.md ├── zigzag-conversion.md ├── zigzag-iterator-ii.md ├── zigzag-iterator.md └── zombie-in-matrix.md └── 高频 ├── anagrams.md ├── char-to-integer.md ├── compare-strings.md ├── count-and-say.md ├── edit-distance-ii.md ├── excel-sheet-column-title.md ├── expression-expand.md ├── expression-tree-build.md ├── first-position-unique-character.md ├── first-unique-character-in-a-string.md ├── implement-strstr.md ├── integer-to-english-words.md ├── integer-to-roman.md ├── is-subsequence.md ├── left-pad.md ├── length-of-last-word.md ├── longest-common-prefix.md ├── longest-common-substring.md ├── longest-palindromic-substring.md ├── longest-substring-with-at-least-k-repeating-characters.md ├── longest-words.md ├── lowercase-to-uppercase-ii.md ├── lowercase-to-uppercase.md ├── lru-cache.md ├── mini-twitter.md ├── mini-yelp.md ├── monochrome-screen.md ├── multiply-strings.md ├── null-terminated-string.md ├── palindrome-number.md ├── parking-lot.md ├── pascals-triangle-ii.md ├── pascals-triangle.md ├── pass-interview.md ├── permutation-index.md ├── plus-one.md ├── prime-factorization.md ├── rat-jump.md ├── read-n-characters-given-read4-ii-call-multiple-times.md ├── readme.md ├── rectangle-area.md ├── reference.md ├── repeated-dna-sequences.md ├── repeated-substring-pattern.md ├── reverse-words-in-a-string.md ├── roman-to-integer.md ├── rotate-string.md ├── shortest-palindrome.md ├── space-replacement.md ├── string-compression.md ├── string-to-integer-atoi.md ├── string-to-integer.md ├── system-longest-file-path.md ├── url-parser.md ├── validate-ip-address.md ├── 数组题 ├── arraylist.md ├── battleships-in-a-board.md ├── best-time-to-buy-and-sell-stock-ii.md ├── best-time-to-buy-and-sell-stock-iii.md ├── best-time-to-buy-and-sell-stock-iv.md ├── best-time-to-buy-and-sell-stock.md ├── can-place-flowers.md ├── candy.md ├── continuous-subarray-sum-ii.md ├── continuous-subarray-sum.md ├── convert-a-number-to-hexadecimal.md ├── first-missing-positive.md ├── flatten-2d-vector.md ├── flip-string-to-monotone-increasing.md ├── generate-arraylist-with-given-size.md ├── implement-deque-by-circulant-array.md ├── implement-stack.md ├── intersection-of-arrays.md ├── k-closest-numbers-in-sorted-array.md ├── max-of-array.md ├── minimum-moves-to-equal-array-elements-ii.md ├── minimum-moves-to-equal-array-elements.md ├── missing-interval.md ├── number-of-airplanes-in-the-sky.md ├── product-of-array-exclude-itself.md ├── remove-duplicates-from-sorted-array-ii.md ├── remove-duplicates-from-sorted-array.md ├── reorder-array-to-construct-the-minimum-number.md ├── rotate-image.md ├── search-a-2d-matrix-ii.md ├── search-in-rotated-sorted-array.md ├── second-max-of-array.md ├── set-matrix-zeroes.md ├── sliding-window-matrix-maximum.md ├── sparse-matrix-multiplication.md ├── spiral-matrix-ii.md ├── spiral-matrix.md ├── subarray-sum-closest.md ├── subarray-sum.md ├── submatrix-sum.md ├── swap-two-integers-in-array.md └── trapping-rain-water.md └── 算法面试必刷100题.xlsx /.gitconfig: -------------------------------------------------------------------------------- 1 | [remote "ninechapter-algorithm"] 2 | url = https://github.com/ninechapter-algorithm/linghu-algorithm-templete.git 3 | fetch = +refs/heads/*:refs/remotes/ninechapter-algorithm/* 4 | [remote "origin"] 5 | url = git@github.com:huangyingw/ninechapter-algorithm_linghu-algorithm-templete.git 6 | fetch = +refs/heads/*:refs/remotes/origin/* 7 | [push] 8 | remote = origin 9 | [gsync] 10 | remote = 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-------------------------------------------------------------------------------- /其他/readme.md: -------------------------------------------------------------------------------- 1 | 2 | -------------------------------------------------------------------------------- /动态规划/backpack-iv.md: -------------------------------------------------------------------------------- 1 | # 背包问题 IV 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack-iv/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出 n 个物品, 以及一个数组, `nums[i]`代表第i个物品的大小, 保证大小均为正数并且没有重复, 正整数 `target` 表示背包的大小, 找到能填满背包的方案数。 5 | `每一个物品可以使用无数次` 6 | ### 样例说明 7 | **样例1** 8 | 9 | ``` 10 | 输入: nums = [2,3,6,7] 和 target = 7 11 | 输出: 2 12 | 解释: 13 | 方案有: 14 | [7] 15 | [2, 2, 3] 16 | ``` 17 | 18 | **样例2** 19 | 20 | ``` 21 | 输入: nums = [2,3,4,5] 和 target = 7 22 | 输出: 3 23 | 解释: 24 | 方案有: 25 | [2, 5] 26 | [3, 4] 27 | [2, 2, 3] 28 | ``` 29 | ### 参考代码 30 | 效率比较高的版本。时间复杂度 $O(n * target)$ 31 | ```python 32 | class Solution: 33 | """ 34 | @param nums: an integer array and all positive numbers, no duplicates 35 | @param target: An integer 36 | @return: An integer 37 | """ 38 | def backPackIV(self, nums, target): 39 | dp = [0] * (target + 1) 40 | dp[0] = 1 41 | 42 | for num in nums: 43 | for i in range(num, target + 1): 44 | dp[i] += dp[i - num] 45 | 46 | return dp[target] 47 | ``` 48 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/backpack-v.md: -------------------------------------------------------------------------------- 1 | # 背包问题 V 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack-v/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出 n 个物品, 以及一个数组, `nums[i]` 代表第i个物品的大小, 保证大小均为正数, 正整数 `target` 表示背包的大小, 找到能填满背包的方案数。 5 | `每一个物品只能使用一次` 6 | ### 样例说明 7 | 给出候选物品集合 `[1,2,3,3,7]` 以及 target `7` 8 | ``` 9 | 结果的集合为: 10 | [7] 11 | [1,3,3] 12 | ``` 13 | 返回 `2` 14 | ### 参考代码 15 | 带一点优化的动态规划。 16 | 17 | - 滚动数组。优化了空间。 18 | - 前缀和。优化了时间。没有必要每次都算到 target 那么多的值。 19 | ```python 20 | class Solution: 21 | """ 22 | @param nums: an integer array and all positive numbers 23 | @param target: An integer 24 | @return: An integer 25 | """ 26 | def backPackV(self, nums, target): 27 | f = [[0] * (target + 1), [0] * (target + 1)] 28 | f[0][0] = 1 29 | n = len(nums) 30 | prefix_sum = 0 31 | for i in range(1, n + 1): 32 | f[i % 2][0] = 1 33 | prefix_sum += nums[i - 1] 34 | for j in range(1, min(target, prefix_sum) + 1): 35 | f[i % 2][j] = f[(i - 1) % 2][j] 36 | if j >= nums[i - 1]: 37 | f[i % 2][j] += f[(i - 1) % 2][j - nums[i - 1]] 38 | 39 | return f[n % 2][target] 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/backpack-vi.md: -------------------------------------------------------------------------------- 1 | # 组合总和 IV 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack-vi/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一个都是正整数的数组 `nums`，其中没有重复的数。从中找出所有的和为 `target` 的组合个数。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | dp求解，dp[i]表示当前和为i的解法个数 9 | ```python 10 | class Solution: 11 | # @param {int[]} nums an integer array and all positive numbers, no duplicates 12 | # @param {int} target an integer 13 | # @return {int} an integer 14 | def backPackVI(self, nums, target): 15 | # Write your code here 16 | dp = [0 for i in xrange(target + 1)] 17 | dp[0] = 1 18 | 19 | for j in xrange(1, target+1): 20 | for a in nums: 21 | if j >= a: 22 | dp[j] += dp[j - a] 23 | 24 | return dp[target] 25 | ``` 26 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/backpack.md: -------------------------------------------------------------------------------- 1 | # 背包问题 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

在n个物品中挑选若干物品装入背包，最多能装多满？假设背包的大小为m，每个物品的大小为A[i]

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: [3,4,8,5], backpack size=10 9 | 输出: 9 10 | 11 | 样例 2: 12 | 输入: [2,3,5,7], backpack size=12 13 | 输出: 12 14 | 15 | ``` 16 | ### 参考代码 17 | 传统的背包问题解法。二维动态规划。 18 | ```python 19 | class Solution: 20 | """ 21 | @param m: An integer m denotes the size of a backpack 22 | @param A: Given n items with size A[i] 23 | @return: The maximum size 24 | """ 25 | def backPack(self, m, A): 26 | n = len(A) 27 | f = [[False] * (m + 1) for _ in range(n + 1)] 28 | 29 | f[0][0] = True 30 | for i in range(1, n + 1): 31 | f[i][0] = True 32 | for j in range(1, m + 1): 33 | if j >= A[i - 1]: 34 | f[i][j] = f[i - 1][j] or f[i - 1][j - A[i - 1]] 35 | else: 36 | f[i][j] = f[i - 1][j] 37 | 38 | for i in range(m, -1, -1): 39 | if f[n][i]: 40 | return i 41 | return 0 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/climbing-stairs-ii.md: -------------------------------------------------------------------------------- 1 | # 爬楼梯 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/climbing-stairs-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 一个小孩爬一个 n 层台阶的楼梯。他可以每次跳 1 步， 2 步 或者 3 步。实现一个方法来统计总共有多少种不同的方式爬到最顶层的台阶。 5 | 6 | ### 样例说明 7 | 8 | ### 参考代码 9 | 本题采用动态规划的思想，每一步的答案可以由上三步转移而来。 10 | 初始状态和状态转移方程如下： 11 | $f_0=1,\ f_1 =1,\ f_2=2,f_n=f_{n-1}+f_{n-2}+f_{n-3}(n \geq 3)$ 12 | ```java 13 | public class Solution { 14 | /** 15 | * @param n an integer 16 | * @return an integer 17 | */ 18 | public int climbStairs2(int n) { 19 | int[] f = new int[n+1]; 20 | f[0] = 1; 21 | for (int i = 0; i <= n; i++) 22 | for (int j = 1; j <= 3; j++) { 23 | if (i >= j) { 24 | f[i] += f[i-j]; 25 | } 26 | } 27 | return f[n]; 28 | } 29 | } 30 | ``` 31 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/coins-in-a-line.md: -------------------------------------------------------------------------------- 1 | # 硬币排成线 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/coins-in-a-line/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 有 `n` 个硬币排成一条线。两个参赛者轮流从右边依次拿走 1 或 2 个硬币，直到没有硬币为止。拿到最后一枚硬币的人获胜。 5 | 6 | 请判定 **先手玩家** 必胜还是必败? 7 | 8 | 若必胜, 返回 `true`, 否则返回 `false`. 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: 1 14 | 输出: true 15 | ``` 16 | 17 | **样例 2:** 18 | 19 | ``` 20 | 输入: 4 21 | 输出: true 22 | 解释: 23 | 先手玩家第一轮拿走一个硬币, 此时还剩三个. 24 | 这时无论后手玩家拿一个还是两个, 下一次先手玩家都可以把剩下的硬币拿完. 25 | ``` 26 | ### 参考代码 27 | 可以证明, 当硬币数目是3的倍数的时候, 先手玩家必败, 否则他必胜. 28 | 29 | 当硬币数目是3的倍数时, 每一轮先手者拿a个, 后手者拿3-a个即可, 后手必胜. 30 | 31 | 若不是3的倍数, 先手者可以拿1或2个, 此时剩余硬币个数就变成了3的倍数. 32 | ```cpp 33 | class Solution { 34 | public: 35 | /** 36 | * @param n: An integer 37 | * @return: A boolean which equals to true if the first player will win 38 | */ 39 | bool firstWillWin(int n) { 40 | return n % 3; 41 | } 42 | }; 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/drop-eggs.md: -------------------------------------------------------------------------------- 1 | # 丢鸡蛋 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/drop-eggs/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 楼有 `n` 层高，鸡蛋若从 *k* 层或以上扔，就会碎。从 *k* 层以下扔，就不会碎。 5 | 6 | 现在给两个鸡蛋，用最少的扔的次数找到 *k*。返回最坏情况下次数。 7 | ### 样例说明 8 | **样例 1：** 9 | ``` 10 | 输入：100 11 | 输出：14 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：10 16 | 输出：4 17 | ``` 18 | ### 参考代码 19 | ```cpp 20 | class Solution { 21 | public: 22 | /** 23 | * @param n an integer 24 | * @return an integer 25 | */ 26 | int dropEggs(int n) { 27 | // Write your code here 28 | long long ans = 0; 29 | int x = 0; 30 | while (ans < n) { 31 | x += 1; 32 | ans += x; 33 | } 34 | return x; 35 | } 36 | }; 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/house-robber.md: -------------------------------------------------------------------------------- 1 | # 打劫房屋 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/house-robber/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 假设你是一个专业的窃贼，准备沿着一条街打劫房屋。每个房子都存放着特定金额的钱。你面临的唯一约束条件是：相邻的房子装着相互联系的防盗系统，且 **当相邻的两个房子同一天被打劫时，该系统会自动报警**。 5 | 6 | 给定一个非负整数列表，表示每个房子中存放的钱， 算一算，如果今晚去打劫，**在不触动报警装置的情况下**, 你最多可以得到多少钱 。 7 | ### 样例说明 8 | **样例 1:** 9 | 10 | ``` 11 | 输入: [3, 8, 4] 12 | 输出: 8 13 | 解释: 仅仅打劫第二个房子. 14 | ``` 15 | 16 | **样例 2:** 17 | 18 | ``` 19 | 输入: [5, 2, 1, 3] 20 | 输出: 8 21 | 解释: 抢第一个和最后一个房子 22 | ``` 23 | ### 参考代码 24 | 使用滚动数组的做法，将每个下标直接 % 3。 25 | 时间复杂度 $O(n)$，空间复杂度 $O(1)$ 26 | ```python 27 | class Solution: 28 | """ 29 | @param A: An array of non-negative integers 30 | @return: The maximum amount of money you can rob tonight 31 | """ 32 | def houseRobber(self, A): 33 | if not A: 34 | return 0 35 | if len(A) <= 2: 36 | return max(A) 37 | 38 | f = [0] * 3 39 | f[0], f[1] = A[0], max(A[0], A[1]) 40 | 41 | for i in range(2, len(A)): 42 | f[i % 3] = max(f[(i - 1) % 3], f[(i - 2) % 3] + A[i]) 43 | 44 | return f[(len(A) - 1) % 3] 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/longest-common-subsequence.md: -------------------------------------------------------------------------------- 1 | # 最长公共子序列 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/longest-common-subsequence/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给出两个字符串，找到最长公共子序列(LCS)，返回LCS的长度。

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: "ABCD" and "EDCA" 9 | 输出: 1 10 | 11 | 解释: 12 | LCS 是 'A' 或 'D' 或 'C' 13 | 14 | 15 | 样例 2: 16 | 输入: "ABCD" and "EACB" 17 | 输出: 2 18 | 19 | 解释: 20 | LCS 是 "AC" 21 | ``` 22 | 23 | ### 参考代码 24 | 使用九章算法强化班和动态规划专题班中讲的匹配型动态规划 25 | ```python 26 | class Solution: 27 | """ 28 | @param A: A string 29 | @param B: A string 30 | @return: The length of longest common subsequence of A and B 31 | """ 32 | def longestCommonSubsequence(self, A, B): 33 | n, m = len(A), len(B) 34 | f = [[0] * (m + 1) for i in range(n + 1)] 35 | for i in range(1, n + 1): 36 | for j in range(1, m + 1): 37 | f[i][j] = max(f[i][j - 1], f[i - 1][j]) 38 | if A[i - 1] == B[j - 1]: 39 | f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1) 40 | return f[n][m] 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /动态规划/maximum-subarray.md: -------------------------------------------------------------------------------- 1 | # 最大子数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/maximum-subarray/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组，找到一个具有最大和的子数组，返回其最大和。 5 | ### 样例说明 6 | **样例1:** 7 | 8 | ``` 9 | 输入：[−2,2,−3,4,−1,2,1,−5,3] 10 | 输出：6 11 | 解释：符合要求的子数组为[4,−1,2,1]，其最大和为 6。 12 | ``` 13 | 14 | **样例2:** 15 | 16 | ``` 17 | 输入：[1,2,3,4] 18 | 输出：10 19 | 解释：符合要求的子数组为[1,2,3,4]，其最大和为 10。 20 | ``` 21 | ### 参考代码 22 | 使用前缀和的方法，计算每个位置为结尾的 subarray 的最大值是多少。 23 | ```python 24 | class Solution: 25 | """ 26 | @param nums: A list of integers 27 | @return: A integer indicate the sum of max subarray 28 | """ 29 | def maxSubArray(self, nums): 30 | #prefix_sum记录前i个数的和，max_Sum记录全局最大值，min_Sum记录前i个数中0-k的最小值 31 | min_sum, max_sum = 0, -sys.maxsize 32 | prefix_sum = 0 33 | 34 | for num in nums: 35 | prefix_sum += num 36 | max_sum = max(max_sum, prefix_sum - min_sum) 37 | min_sum = min(min_sum, prefix_sum) 38 | 39 | return max_sum 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/宽度优先搜索/best-time-to-buy-and-sell-stock-ii.md: -------------------------------------------------------------------------------- 1 | # 买卖股票的最佳时机 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/best-time-to-buy-and-sell-stock-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个数组 `prices` 表示一支股票每天的价格. 5 | 6 | 你可以完成任意次数的交易, 不过你不能同时参与多个交易 (也就是说, 如果你已经持有这支股票, 在再次购买之前, 你必须先卖掉它). 7 | 8 | 设计一个算法求出最大的利润. 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: [2, 1, 2, 0, 1] 14 | 输出: 2 15 | 解释: 16 | 1. 在第 2 天以 1 的价格买入, 然后在第 3 天以 2 的价格卖出, 利润 1 17 | 2. 在第 4 天以 0 的价格买入, 然后在第 5 天以 1 的价格卖出, 利润 1 18 | 总利润 2. 19 | ``` 20 | 21 | **样例 2:** 22 | 23 | ``` 24 | 输入: [4, 3, 2, 1] 25 | 输出: 0 26 | 解释: 不进行任何交易, 利润为0. 27 | ``` 28 | ### 参考代码 29 | 贪心法: 只要相邻的两天股票的价格是上升的, 我们就进行一次交易, 获得一定利润. 30 | 31 | 这样的策略不一定是最小的交易次数, 但是一定会得到最大的利润. 32 | ```java 33 | public class Solution { 34 | public int maxProfit(int[] prices) { 35 | int profit = 0; 36 | for (int i = 0; i < prices.length - 1; i++) { 37 | int diff = prices[i + 1] - prices[i]; 38 | if (diff > 0) { 39 | profit += diff; 40 | } 41 | } 42 | return profit; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/宽度优先搜索/path-sum.md: -------------------------------------------------------------------------------- 1 | # 路径和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/path-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定二叉树和求和，确定树是否具有根到叶路径，使得沿路径的所有值相加等于给定的总和。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | 使用深度搜索的方法，当搜索到叶子节点时，比较与目标和的值，若相同，则结果为真，若搜索完毕没有出现目标和，返回假 9 | ```java 10 | /** 11 | * Definition of TreeNode: 12 | * public class TreeNode { 13 | * public int val; 14 | * public TreeNode left, right; 15 | * public TreeNode(int val) { 16 | * this.val = val; 17 | * this.left = this.right = null; 18 | * } 19 | * } 20 | */ 21 | 22 | public class Solution { 23 | /** 24 | * @param root: the tree 25 | * @param sum: the sum 26 | * @return: if the tree has a root-to-leaf path 27 | */ 28 | public boolean pathSum(TreeNode root, int sum) { 29 | // Write your code here. 30 | if (root == null) return false; 31 | else if (root.val == sum && root.left == null && root.right == null) return true; 32 | else return pathSum(root.left, sum - root.val) || pathSum(root.right, sum - root.val); 33 | } 34 | } 35 | ``` 36 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/深度优先搜索/combination-sum-iv.md: -------------------------------------------------------------------------------- 1 | # 组合总和 IV 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/combination-sum-iv/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一个都是正整数的数组 `nums`，其中没有重复的数。从中找出所有的和为 `target` 的组合个数。 5 | ### 样例说明 6 | **样例1** 7 | 8 | ``` 9 | 输入: nums = [1, 2, 4] 和 target = 4 10 | 输出: 6 11 | 解释: 12 | 可能的所有组合有： 13 | [1, 1, 1, 1] 14 | [1, 1, 2] 15 | [1, 2, 1] 16 | [2, 1, 1] 17 | [2, 2] 18 | [4] 19 | ``` 20 | 21 | **样例2** 22 | 23 | ``` 24 | 输入: nums = [1, 2] 和 target = 4 25 | 输出: 5 26 | 解释: 27 | 可能的所有组合有： 28 | [1, 1, 1, 1] 29 | [1, 1, 2] 30 | [1, 2, 1] 31 | [2, 1, 1] 32 | [2, 2] 33 | ``` 34 | ### 参考代码 35 | ```cpp 36 | class Solution { 37 | public: 38 | int combinationSum4(vector& nums, int target) { 39 | int n = nums.size(); 40 | int f[target+1]; 41 | memset(f, 0, sizeof(f)); 42 | 43 | f[0] = 1; 44 | for (int i = 1; i <= target; i++) { 45 | for (int j = 0; j < n; j++) 46 | if (i - nums[j] >= 0){ 47 | f[i] += f[i-nums[j]]; 48 | } 49 | } 50 | return f[target]; 51 | } 52 | }; 53 | ``` 54 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/深度优先搜索/flatten-nested-iterate.md: -------------------------------------------------------------------------------- 1 | # flatten nested iterate 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/flatten-nested-iterate/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 1753 5 | ### 样例说明 6 | 1753 7 | ### 参考代码 8 | ```python 9 | def flattern(A): 10 | result = [] 11 | sta = list() 12 | for ele in reversed(A): 13 | sta.append(ele) 14 | while len(sta) > 0: 15 | ele = sta[-1] 16 | sta.pop() 17 | if isinstance(ele, int): 18 | result.append(ele) 19 | else: 20 | for sub_ele in reversed(ele): 21 | sta.append(sub_ele) 22 | return result 23 | ``` 24 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/深度优先搜索/maximum-subarray.md: -------------------------------------------------------------------------------- 1 | # 最大子数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/maximum-subarray/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组，找到一个具有最大和的子数组，返回其最大和。 5 | ### 样例说明 6 | **样例1:** 7 | 8 | ``` 9 | 输入：[−2,2,−3,4,−1,2,1,−5,3] 10 | 输出：6 11 | 解释：符合要求的子数组为[4,−1,2,1]，其最大和为 6。 12 | ``` 13 | 14 | **样例2:** 15 | 16 | ``` 17 | 输入：[1,2,3,4] 18 | 输出：10 19 | 解释：符合要求的子数组为[1,2,3,4]，其最大和为 10。 20 | ``` 21 | ### 参考代码 22 | 使用前缀和的方法，计算每个位置为结尾的 subarray 的最大值是多少。 23 | ```python 24 | class Solution: 25 | """ 26 | @param nums: A list of integers 27 | @return: A integer indicate the sum of max subarray 28 | """ 29 | def maxSubArray(self, nums): 30 | #prefix_sum记录前i个数的和，max_Sum记录全局最大值，min_Sum记录前i个数中0-k的最小值 31 | min_sum, max_sum = 0, -sys.maxsize 32 | prefix_sum = 0 33 | 34 | for num in nums: 35 | prefix_sum += num 36 | max_sum = max(max_sum, prefix_sum - min_sum) 37 | min_sum = min(min_sum, prefix_sum) 38 | 39 | return max_sum 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/深度优先搜索/next-permutation.md: -------------------------------------------------------------------------------- 1 | # 下一个排列 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/next-permutation/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组来表示排列，找出其之后的一个排列。 5 | ### 样例说明 6 | 例1: 7 | ``` 8 | 输入:[1] 9 | 输出:[1] 10 | ``` 11 | 12 | 13 | 例2: 14 | ``` 15 | 输入:[1,3,2,3] 16 | 输出:[1,3,3,2] 17 | ``` 18 | 19 | 例3: 20 | ``` 21 | 输入:[4,3,2,1] 22 | 输出:[1,2,3,4] 23 | ``` 24 | 25 | ### 参考代码 26 | 从最后一个位置开始，找到一个上升点，上升点之前的无需改动。 27 | 然后，翻转上升点之后的降序。 28 | 在降序里，找到第一个比上升点大的，交换位置。 29 | ```python 30 | class Solution: 31 | # @param num : a list of integer 32 | # @return : a list of integer 33 | def nextPermutation(self, num): 34 | # write your code here 35 | for i in range(len(num)-2, -1, -1): 36 | if num[i] < num[i+1]: 37 | break 38 | else: 39 | num.reverse() 40 | return num 41 | for j in range(len(num)-1, i, -1): 42 | if num[j] > num[i]: 43 | num[i], num[j] = num[j], num[i] 44 | break 45 | for j in range(0, (len(num) - i)//2): 46 | num[i+j+1], num[len(num)-j-1] = num[len(num)-j-1], num[i+j+1] 47 | return num 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/深度优先搜索/prime-factorization.md: -------------------------------------------------------------------------------- 1 | # 分解质因数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/prime-factorization/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个整数分解为若干质因数之乘积。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：10 9 | 输出：[2, 5] 10 | ``` 11 | **样例 2：** 12 | ``` 13 | 输入：660 14 | 输出：[2, 2, 3, 5, 11] 15 | ``` 16 | ### 参考代码 17 | ```java 18 | public class Solution { 19 | /** 20 | * @param num an integer 21 | * @return an integer array 22 | */ 23 | public List primeFactorization(int num) { 24 | List factors = new ArrayList(); 25 | 26 | for (int i = 2; i * i <= num; i++) { 27 | while (num % i == 0) { 28 | num = num / i; 29 | factors.add(i); 30 | } 31 | } 32 | 33 | if (num != 1) { 34 | factors.add(num); 35 | } 36 | 37 | return factors; 38 | } 39 | } 40 | 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /图论/深度优先搜索/subsets-ii.md: -------------------------------------------------------------------------------- 1 | # 子集 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/subsets-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个可能具有重复数字的列表，返回其所有可能的子集。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：[0] 9 | 输出： 10 | [ 11 | [], 12 | [0] 13 | ] 14 | ``` 15 | **样例 2：** 16 | ``` 17 | 输入：[1,2,2] 18 | 输出： 19 | [ 20 | [2], 21 | [1], 22 | [1,2,2], 23 | [2,2], 24 | [1,2], 25 | [] 26 | ] 27 | ``` 28 | ### 参考代码 29 | 九章算法班中的方法 30 | ```python 31 | class Solution: 32 | """ 33 | @param nums: A set of numbers. 34 | @return: A list of lists. All valid subsets. 35 | """ 36 | def subsetsWithDup(self, nums): 37 | nums = sorted(nums) 38 | subsets = [] 39 | self.dfs(nums, 0, [], subsets) 40 | return subsets 41 | 42 | def dfs(self, nums, index, subset, subsets): 43 | subsets.append(list(subset)) 44 | 45 | for i in range(index, len(nums)): 46 | if i > index and nums[i] == nums[i - 1]: 47 | continue 48 | subset.append(nums[i]) 49 | self.dfs(nums, i + 1, subset, subsets) 50 | subset.pop() 51 | ``` 52 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/kth-largest-element-ii.md: -------------------------------------------------------------------------------- 1 | # 第K大的元素 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/kth-largest-element-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 找到数组中第K大的元素，N远大于K。请注意你需要找的是数组排序后的第 k 个最大的元素，而不是第 k 个不同的元素 5 | ### 样例说明 6 | **例1:** 7 | ``` 8 | 输入:[9,3,2,4,8],3 9 | 输出:4 10 | 11 | ``` 12 | **例2:** 13 | ``` 14 | 输入:[1,2,3,4,5,6,8,9,10,7],10 15 | 输出:1 16 | 17 | ``` 18 | 19 | ### 参考代码 20 | 利用优先队列求第k大元素 21 | ```java 22 | class Solution { 23 | /** 24 | * @param nums an integer unsorted array 25 | * @param k an integer from 1 to n 26 | * @return the kth largest element 27 | */ 28 | public int kthLargestElement2(int[] nums, int k) { 29 | // Write your code here 30 | PriorityQueue q = new PriorityQueue(k); 31 | for (int num : nums) { 32 | q.offer(num); 33 | if (q.size() > k) { 34 | q.poll(); 35 | } 36 | } 37 | return q.peek(); 38 | } 39 | }; 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/moving-average-from-data-stream.md: -------------------------------------------------------------------------------- 1 | # 数据流滑动窗口平均值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/moving-average-from-data-stream/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一串整数流和窗口大小，计算滑动窗口中所有整数的平均值。 5 | ### 样例说明 6 | **样例1 :** 7 | ``` 8 | MovingAverage m = new MovingAverage(3); 9 | m.next(1) = 1 // 返回 1.00000 10 | m.next(10) = (1 + 10) / 2 // 返回 5.50000 11 | m.next(3) = (1 + 10 + 3) / 3 // 返回 4.66667 12 | m.next(5) = (10 + 3 + 5) / 3 // 返回 6.00000 13 | ``` 14 | ### 参考代码 15 | 使用队列 16 | ```java 17 | public class MovingAverage { 18 | 19 | private Queue que; 20 | private double sum = 0; 21 | private int size; 22 | 23 | /** Initialize your data structure here. */ 24 | public MovingAverage(int size) { 25 | que = new LinkedList(); 26 | this.size = size; 27 | } 28 | 29 | public double next(int val) { 30 | // Write your code here 31 | sum += val; 32 | if (que.size() == size) { 33 | sum = sum - que.poll(); 34 | } 35 | que.offer(val); 36 | return sum / que.size(); 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/print-numbers-by-recursion.md: -------------------------------------------------------------------------------- 1 | # 用递归打印数字 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/print-numbers-by-recursion/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用递归的方法找到从1到最大的N位整数。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入 : N = 1 9 | 输出 :[1,2,3,4,5,6,7,8,9] 10 | ``` 11 | **样例 2:** 12 | ``` 13 | 输入 : N = 2 14 | 输出 :[[1,2,3,4,5,6,7,8,9,10,11,12,...,99] 15 | ``` 16 | 17 | ### 参考代码 18 | ```java 19 | public class Solution { 20 | /** 21 | * @param n: An integer. 22 | * return : An array storing 1 to the largest number with n digits. 23 | */ 24 | public List numbersByRecursion(int n) { 25 | // write your code here 26 | ArrayList res = new ArrayList<>(); 27 | num(n, 0, res); 28 | return res; 29 | } 30 | 31 | public void num(int n, int ans,ArrayList res){ 32 | 33 | if(n == 0){ 34 | if(ans > 0){ 35 | res.add(ans); 36 | } 37 | return; 38 | } 39 | 40 | int i; 41 | for(i = 0; i <= 9; i++){ 42 | num(n - 1, ans * 10 + i, res); 43 | } 44 | 45 | } 46 | } 47 | ``` 48 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/⼆叉搜索树/binary-tree-leaf-sum.md: -------------------------------------------------------------------------------- 1 | # 二叉树叶子节点之和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/binary-tree-leaf-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算二叉树的叶子节点之和 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入：{1,2,3,4} 9 | 输出：7 10 | 解释： 11 | 1 12 | / \ 13 | 2 3 14 | / 15 | 4 16 | 3+4=7 17 | ``` 18 | **样例 2:** 19 | ``` 20 | 输入：{1,#,3} 21 | 输出：3 22 | 解释： 23 | 1 24 | \ 25 | 3 26 | ``` 27 | ### 参考代码 28 | 递归求和即可 29 | ```python 30 | """ 31 | Definition of TreeNode: 32 | class TreeNode: 33 | def __init__(self, val): 34 | this.val = val 35 | this.left, this.right = None, None 36 | """ 37 | class Solution: 38 | # @param {TreeNode} root the root of the binary tree 39 | # @return {int} an integer 40 | def leafSum(self, root): 41 | # Write your code here 42 | p = [] 43 | self.dfs(root, p) 44 | return sum(p) 45 | def dfs(self, root, p): 46 | if root is None: 47 | return 48 | if root.left is None and root.right is None: 49 | p.append(root.val) 50 | self.dfs(root.left, p) 51 | self.dfs(root.right, p) 52 | ``` 53 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/⼆叉搜索树/count-1-in-binary.md: -------------------------------------------------------------------------------- 1 | # 二进制中有多少个1 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-1-in-binary/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算在一个 32 位的整数的二进制表示中有多少个 `1`。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：32 9 | 输出：1 10 | 解释： 11 | 32(100000)，返回 1。 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：5 16 | 输出：2 17 | 解释： 18 | 5(101)，返回 2。 19 | ``` 20 | ### 参考代码 21 | ```java 22 | public class Solution { 23 | /** 24 | * @param num: an integer 25 | * @return: an integer, the number of ones in num 26 | */ 27 | public int countOnes(int num) { 28 | int count = 0; 29 | while (num != 0) { 30 | num = num & (num - 1); 31 | count++; 32 | } 33 | return count; 34 | } 35 | }; 36 | 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/二叉树分治/binary-tree-leaf-sum.md: -------------------------------------------------------------------------------- 1 | # 二叉树叶子节点之和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/binary-tree-leaf-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算二叉树的叶子节点之和 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入：{1,2,3,4} 9 | 输出：7 10 | 解释： 11 | 1 12 | / \ 13 | 2 3 14 | / 15 | 4 16 | 3+4=7 17 | ``` 18 | **样例 2:** 19 | ``` 20 | 输入：{1,#,3} 21 | 输出：3 22 | 解释： 23 | 1 24 | \ 25 | 3 26 | ``` 27 | ### 参考代码 28 | 递归求和即可 29 | ```python 30 | """ 31 | Definition of TreeNode: 32 | class TreeNode: 33 | def __init__(self, val): 34 | this.val = val 35 | this.left, this.right = None, None 36 | """ 37 | class Solution: 38 | # @param {TreeNode} root the root of the binary tree 39 | # @return {int} an integer 40 | def leafSum(self, root): 41 | # Write your code here 42 | p = [] 43 | self.dfs(root, p) 44 | return sum(p) 45 | def dfs(self, root, p): 46 | if root is None: 47 | return 48 | if root.left is None and root.right is None: 49 | p.append(root.val) 50 | self.dfs(root.left, p) 51 | self.dfs(root.right, p) 52 | ``` 53 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/哈希表/anagrams.md: -------------------------------------------------------------------------------- 1 | # 乱序字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/anagrams/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给出一个字符串数组S，找到其中所有的乱序字符串(Anagram)。如果一个字符串是乱序字符串，那么他存在一个字母集合相同，但顺序不同的字符串也在S中。

5 | ### 样例说明 6 | **样例1:** 7 | ``` 8 | 输入:["lint", "intl", "inlt", "code"] 9 | 输出:["lint", "inlt", "intl"] 10 | ``` 11 | 12 | **样例 2:** 13 | ``` 14 | 输入:["ab", "ba", "cd", "dc", "e"] 15 | 输出: ["ab", "ba", "cd", "dc"] 16 | ``` 17 | ### 参考代码 18 | Given an array of strings, return all groups of strings that are anagrams. 19 | 20 | Note: All inputs will be in lower-case 21 | ```python 22 | class Solution: 23 | # @param strs: A list of strings 24 | # @return: A list of strings 25 | def anagrams(self, strs): 26 | # write your code here 27 | dict = {} 28 | for word in strs: 29 | sortedword = ''.join(sorted(word)) 30 | dict[sortedword] = [word] if sortedword not in dict else dict[sortedword] + [word] 31 | res = [] 32 | for item in dict: 33 | if len(dict[item]) >= 2: 34 | res += dict[item] 35 | return res 36 | 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/哈希表/count-characters.md: -------------------------------------------------------------------------------- 1 | # 字符计数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-characters/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 对一个字符串中的字符进行计数, 返回一个hashmap, key为字符, value是这个字符出现的次数. 5 | ### 样例说明 6 | 7 | 例1: 8 | ``` 9 | 输入: 10 | str = "abca" 11 | 12 | 输出: 13 | { 14 | "a": 2, 15 | "b": 1, 16 | "c": 1 17 | } 18 | 19 | ``` 20 | 例2: 21 | ``` 22 | 输入: 23 | str = "ab" 24 | 25 | 输出: 26 | { 27 | "a": 1, 28 | "b": 1 29 | } 30 | 31 | ``` 32 | 33 | ### 参考代码 34 | 利用map统计即可 35 | ```java 36 | public class Solution { 37 | /* 38 | * @param : a string 39 | * @return: a hash map 40 | */ 41 | public Map countCharacters(String str) { 42 | // Write your code here 43 | Map map = new HashMap(); 44 | for (char c : str.toCharArray()) { 45 | if (!map.containsKey(c)) { 46 | map.put(c, 1); 47 | } else { 48 | map.put(c, map.get(c) + 1); 49 | } 50 | } 51 | return map; 52 | } 53 | }; 54 | ``` 55 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/哈希表/majority-number.md: -------------------------------------------------------------------------------- 1 | # 主元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/majority-number/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个整型数组，找出主元素，它在数组中的出现次数严格大于数组元素个数的二分之一。


5 | ### 样例说明 6 | 7 | ### 参考代码 8 |

    

9 | ```java 10 | public class Solution { 11 | /** 12 | * @param nums: a list of integers 13 | * @return: find a majority number 14 | */ 15 | public int majorityNumber(ArrayList nums) { 16 | //由于主元素在数组中个数严格大于1/2，所以假设当前元素就是主元素，用count记录当前元素个数与其他元素个数的差值， 17 | //candidate记录当前元素是什么，最后count一定为正数并且candidate就是主元素 18 | int count = 0, candidate = -1; 19 | for (int i = 0; i < nums.size(); i++) { 20 | if (count == 0) { 21 | candidate = nums.get(i); 22 | count = 1; 23 | } else if (candidate == nums.get(i)) { 24 | count++; 25 | } else { 26 | count--; 27 | } 28 | } 29 | return candidate; 30 | } 31 | } 32 | ``` 33 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/哈希表/name-deduplication.md: -------------------------------------------------------------------------------- 1 | # 姓名去重 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/name-deduplication/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一串名字，将他们去重之后返回。两个名字重复是说在忽略大小写的情况下是一样的。 5 | ### 样例说明 6 | 例1: 7 | ``` 8 | 输入:["James", "james", "Bill Gates", "bill Gates", "Hello World", "HELLO WORLD", "Helloworld"] 9 | 10 | 11 | 输出:["james", "bill gates", "hello world", "helloworld"] 12 | ``` 13 | 例2: 14 | ``` 15 | 输入:["cmy","Cmy"] 16 | 17 | 输出:["cmy"] 18 | ``` 19 | ### 参考代码 20 | 利用map去重 21 | ```cpp 22 | #include 23 | class Solution { 24 | public: 25 | /** 26 | * @param names a string array 27 | * @return a string array 28 | */ 29 | vector nameDeduplication(vector& names) { 30 | // Write your code here 31 | map mp; 32 | vector result; 33 | for (auto& str : names) { 34 | std::transform(str.begin(), str.end(), str.begin(), ::tolower); 35 | if (mp.find(str) == mp.end()) { 36 | mp[str] = true; 37 | result.push_back(str); 38 | } 39 | } 40 | return result; 41 | } 42 | }; 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/堆/kth-largest-element-ii.md: -------------------------------------------------------------------------------- 1 | # 第K大的元素 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/kth-largest-element-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 找到数组中第K大的元素，N远大于K。请注意你需要找的是数组排序后的第 k 个最大的元素，而不是第 k 个不同的元素 5 | ### 样例说明 6 | **例1:** 7 | ``` 8 | 输入:[9,3,2,4,8],3 9 | 输出:4 10 | 11 | ``` 12 | **例2:** 13 | ``` 14 | 输入:[1,2,3,4,5,6,8,9,10,7],10 15 | 输出:1 16 | 17 | ``` 18 | 19 | ### 参考代码 20 | 利用优先队列求第k大元素 21 | ```java 22 | class Solution { 23 | /** 24 | * @param nums an integer unsorted array 25 | * @param k an integer from 1 to n 26 | * @return the kth largest element 27 | */ 28 | public int kthLargestElement2(int[] nums, int k) { 29 | // Write your code here 30 | PriorityQueue q = new PriorityQueue(k); 31 | for (int num : nums) { 32 | q.offer(num); 33 | if (q.size() > k) { 34 | q.poll(); 35 | } 36 | } 37 | return q.peek(); 38 | } 39 | }; 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/堆/ugly-number-ii.md: -------------------------------------------------------------------------------- 1 | # 丑数 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/ugly-number-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 设计一个算法，找出只含素因子`2`，`3`，`5` 的第 *n* 小的数。 5 | 6 | 符合条件的数如：`1, 2, 3, 4, 5, 6, 8, 9, 10, 12...` 7 | 8 | ### 样例说明 9 | **样例 1：** 10 | ``` 11 | 输入：9 12 | 输出：10 13 | ``` 14 | **样例 2：** 15 | ``` 16 | 输入：1 17 | 输出：1 18 | ``` 19 | ### 参考代码 20 | 使用九章算法班中讲的 heap 21 | 时间复杂度 $O(nlogn)$ 22 | ```python 23 | import heapq 24 | 25 | class Solution: 26 | """ 27 | @param {int} n an integer. 28 | @return {int} the nth prime number as description. 29 | """ 30 | def nthUglyNumber(self, n): 31 | heap = [1] 32 | visited = set([1]) 33 | 34 | val = None 35 | for i in range(n): 36 | val = heapq.heappop(heap) 37 | for factor in [2, 3, 5]: 38 | if val * factor not in visited: 39 | visited.add(val * factor) 40 | heapq.heappush(heap, val * factor) 41 | 42 | return val 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/栈/simplify-path.md: -------------------------------------------------------------------------------- 1 | # 简化路径 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/simplify-path/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个文件的绝对路径(Unix-style)，请进行路径简化。 5 | 6 | Unix中, `.` 表示当前目录, `..` 表示父目录。 7 | 8 | 结果必须以 `/` 开头，并且两个目录名之间有且只有一个 `/`。最后一个目录名(如果存在)后不能出现 `/` 。你需要保证结果是正确表示路径的最短的字符串。 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: "/home/" 14 | 输出: "/home" 15 | ``` 16 | 17 | **样例 2:** 18 | 19 | ``` 20 | 输入: "/a/./../../c/" 21 | 输出: "/c" 22 | 解释: "/" 没有上级目录, "/../" 的结果就是 "/". 23 | ``` 24 | ### 参考代码 25 | 用栈处理即可. 26 | 27 | 将原字符串以 `'/'` 分隔, 然后遍历: 28 | 29 | - 遇到正常的目录名, 入栈 30 | - 遇到 `'.'` 或 空名称 (对应 `"//"`) 则忽略 31 | - 遇到 `".."` 则从栈顶弹出一个元素 (如果栈为空则不弹栈, 对应 `"/../"`) 32 | 33 | 最后将栈中的元素以 `'/'` 连接得到结果. 34 | ```python 35 | class Solution: 36 | """ 37 | @param path: the original path 38 | @return: the simplified path 39 | """ 40 | def simplifyPath(self, path): 41 | path = path.split('/') 42 | stack = [] 43 | for i in path: 44 | if i == '..': 45 | if len(stack): 46 | stack.pop() 47 | elif i != '.' and i != '': 48 | stack.append(i) 49 | return '/' + '/'.join(stack) 50 | ``` 51 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/convert-linked-list-to-array-list.md: -------------------------------------------------------------------------------- 1 | # 链表转数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/convert-linked-list-to-array-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个链表转换为一个数组。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 1->2->3->null 9 | 输出: [1,2,3] 10 | ``` 11 | **样例 2:** 12 | ``` 13 | 输入: 3->5->8->null 14 | 输出: [3,5,8] 15 | ``` 16 | 17 | 18 | ### 参考代码 19 | ```python 20 | """ 21 | Definition of ListNode 22 | class ListNode(object): 23 | 24 | def __init__(self, val, next=None): 25 | self.val = val 26 | self.next = next 27 | """ 28 | class Solution: 29 | """ 30 | @param {ListNode} head the first node of the linked list. 31 | @return {int[]} an integer list 32 | """ 33 | def toArrayList(self, head): 34 | # Write your code here 35 | r = [] 36 | #遍历链表，并将每个节点的val存储在数组中 37 | while head is not None: 38 | r.append(head.val) 39 | head = head.next 40 | return r 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/count-linked-list-nodes.md: -------------------------------------------------------------------------------- 1 | # 链表节点计数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-linked-list-nodes/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算链表中有多少个节点. 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 1->3->5->null 9 | 输出: 3 10 | 11 | 样例解释: 12 | 返回链表中结点个数，也就是链表的长度. 13 | 14 | 样例 2: 15 | 输入: null 16 | 输出: 0 17 | 18 | 样例解释: 19 | 空链表长度为0 20 | 21 | ``` 22 | ### 参考代码 23 | 简单模拟 24 | ```cpp 25 | /** 26 | * Definition of ListNode 27 | * class ListNode { 28 | * public: 29 | * int val; 30 | * ListNode *next; 31 | * ListNode(int val) { 32 | * this->val = val; 33 | * this->next = NULL; 34 | * } 35 | * } 36 | */ 37 | class Solution { 38 | public: 39 | /** 40 | * @param head the first node of linked list. 41 | * @return an integer 42 | */ 43 | int countNodes(ListNode *head) { 44 | // Write your code here 45 | int cnt = 0; 46 | while (head != NULL) { 47 | cnt ++; 48 | head = head->next; 49 | } 50 | return cnt; 51 | } 52 | }; 53 | ``` 54 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/delete-node-in-the-middle-of-singly-linked-list.md: -------------------------------------------------------------------------------- 1 | # 在O(1)时间复杂度删除链表节点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/delete-node-in-the-middle-of-singly-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个单链表中的一个等待被删除的节点(非表头或表尾)。请在在 O(1) 时间复杂度删除该链表节点。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | ```python 9 | """ 10 | Definition of ListNode 11 | class ListNode(object): 12 | 13 | def __init__(self, val, next=None): 14 | self.val = val 15 | self.next = next 16 | """ 17 | class Solution: 18 | # @param node: the node in the list should be deleted 19 | # @return: nothing 20 | def deleteNode(self, node): 21 | # write your code here 22 | if node is None or node.next is None: 23 | return 24 | next = node.next; 25 | node.val = next.val 26 | node.next = next.next 27 | return 28 | ``` 29 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/find-node-in-linked-list.md: -------------------------------------------------------------------------------- 1 | # 在链表中找节点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/find-node-in-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在链表中找值为 value 的节点，如果没有的话，返回空(null)。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 1->2->3 and value = 3 9 | 输出: 最后一个结点 10 | ``` 11 | **样例 2:** 12 | ``` 13 | 输入: 1->2->3 and value = 4 14 | 输出: null 15 | ``` 16 | ### 参考代码 17 | 简单模拟，复杂度O(n) 18 | ```java 19 | /** 20 | * Definition for ListNode 21 | * public class ListNode { 22 | * int val; 23 | * ListNode next; 24 | * ListNode(int x) { 25 | * val = x; 26 | * next = null; 27 | * } 28 | * } 29 | */ 30 | public class Solution { 31 | /** 32 | * @param head: the head of linked list. 33 | * @param val: an integer 34 | * @return: a linked node or null 35 | */ 36 | public ListNode findNode(ListNode head, int val) { 37 | for (ListNode node = head; node != null; node = node.next) { 38 | if (node.val == val) { 39 | return node; 40 | } 41 | } 42 | return null; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/insert-node-in-sorted-linked-list.md: -------------------------------------------------------------------------------- 1 | # 在排序链表中插入一个节点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/insert-node-in-sorted-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在链表中插入一个节点。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：head = 1->4->6->8->null, val = 5 9 | 输出：1->4->5->6->8->null 10 | ``` 11 | **样例 2：** 12 | ``` 13 | 输入：head = 1->null, val = 2 14 | 输出：1->2->null 15 | ``` 16 | ### 参考代码 17 | ---------------------------------------------------------- 18 | ```python 19 | """ 20 | Definition of ListNode 21 | class ListNode(object): 22 | 23 | def __init__(self, val, next=None): 24 | self.val = val 25 | self.next = next 26 | """ 27 | 28 | class Solution: 29 | # @param head, a ListNode 30 | # @param val, an integer 31 | # @return the head of new linked list 32 | def insertNode(self, head, val): 33 | # Write your code here 34 | dummy = ListNode(0, head) 35 | p = dummy 36 | while p.next and p.next.val < val: 37 | p = p.next 38 | node = ListNode(val, p.next) 39 | p.next = node 40 | return dummy.next 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/insertion-sort-list.md: -------------------------------------------------------------------------------- 1 | # 链表插入排序 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/insertion-sort-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用插入排序对链表排序 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 0->null 9 | 输出: 0->null 10 | 11 | 12 | 样例 2: 13 | 输入: 1->3->2->0->null 14 | 输出 :0->1->2->3->null 15 | 16 | 17 | ``` 18 | 19 | ### 参考代码 20 | Sort a linked list using insertion sort. 21 | 22 | 与数组普通插入排序一样的做法。 23 | ```java 24 | public class Solution { 25 | public ListNode insertionSortList(ListNode head) { 26 | ListNode dummy = new ListNode(0); 27 | // 这个dummy的作用是，把head开头的链表一个个的插入到dummy开头的链表里 28 | // 所以这里不需要dummy.next = head; 29 | 30 | while (head != null) { 31 | ListNode node = dummy; 32 | while (node.next != null && node.next.val < head.val) { 33 | node = node.next; 34 | } 35 | ListNode temp = head.next; 36 | head.next = node.next; 37 | node.next = head; 38 | head = temp; 39 | } 40 | 41 | return dummy.next; 42 | } 43 | } 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/middle-of-linked-list.md: -------------------------------------------------------------------------------- 1 | # 链表的中点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/middle-of-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 找链表的中点。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 1->2->3 9 | 输出: 2 10 | 样例解释: 返回中间节点的值 11 | ``` 12 | **样例 2:** 13 | ``` 14 | 输入: 1->2 15 | 输出: 1 16 | 样例解释: 如果长度是偶数，则返回中间偏左的节点的值。 17 | ``` 18 | ### 参考代码 19 | 这里使用快慢指针的方法来寻找终点。 20 | 可以确保时间复杂度是 O(n/2) 21 | 22 | 快指针每次两步，慢指针每次一步。 23 | 快指针到尾的时候，慢指针就是中点。 24 | ```python 25 | """ 26 | Definition of ListNode 27 | class ListNode(object): 28 | 29 | def __init__(self, val, next=None): 30 | self.val = val 31 | self.next = next 32 | """ 33 | 34 | class Solution: 35 | # @param head: the head of linked list. 36 | # @return: a middle node of the linked list 37 | def middleNode(self, head): 38 | # Write your code here 39 | if head is None: 40 | return None 41 | slow = head 42 | fast = slow.next 43 | while fast is not None and fast.next is not None: 44 | slow = slow.next 45 | fast = fast.next.next 46 | 47 | return slow 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/remove-duplicates-from-sorted-list.md: -------------------------------------------------------------------------------- 1 | # 删除排序链表中的重复元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/remove-duplicates-from-sorted-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个排序链表，删除所有重复的元素每个元素只留下一个。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: null 9 | 输出: null 10 | 11 | 12 | 样例 2: 13 | 输入: 1->1->2->null 14 | 输出: 1->2->null 15 | 16 | 样例 3: 17 | 输入: 1->1->2->3->3->null 18 | 输出: 1->2->3->null 19 | 20 | 21 | ``` 22 | ### 参考代码 23 | Given a sorted linked list, delete all duplicates such that each element appear only once. 24 | 25 | For example, 26 | Given 1->1->2, return 1->2. 27 | Given 1->1->2->3->3, return 1->2->3. 28 | 29 | 跳过相同的元素，指向再后一个元素。 30 | ```java 31 | public class Solution { 32 | public ListNode deleteDuplicates(ListNode head) { 33 | if (head == null) { 34 | return null; 35 | } 36 | 37 | ListNode node = head; 38 | while (node.next != null) { 39 | if (node.val == node.next.val) { 40 | node.next = node.next.next; 41 | } else { 42 | node = node.next; 43 | } 44 | } 45 | return head; 46 | } 47 | } 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /数据结构/链表/reverse-linked-list.md: -------------------------------------------------------------------------------- 1 | # 翻转链表 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reverse-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

翻转一个链表

5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: 1->2->3->null 10 | 输出: 3->2->1->null 11 | ``` 12 | 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: 1->2->3->4->null 17 | 输出: 4->3->2->1->null 18 | ``` 19 | ### 参考代码 20 | http://www.lintcode.com/zh-cn/problem/reverse-linked-list/ 21 | ```python 22 | class Solution: 23 | 24 | def reverse(self, head): 25 | #curt表示前继节点 26 | curt = None 27 | while head != None: 28 | #temp记录下一个节点，head是当前节点 29 | temp = head.next 30 | head.next = curt 31 | curt = head 32 | head = temp 33 | return curt 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/二分法/closest-number-in-sorted-array.md: -------------------------------------------------------------------------------- 1 | # 排序数组中最接近元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/closest-number-in-sorted-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在一个排好序的数组 *A* 中找到 *i* 使得 *A[i]* 最接近 *target* 5 | ### 样例说明 6 | 样例1： 7 | ``` 8 | 输入：[1, 2, 3] ，target = 2 9 | 输出： 1. 10 | ``` 11 | 样例2： 12 | ``` 13 | 输入：[1, 4, 6]， target = 3 14 | 输出： 1. 15 | ``` 16 | 样例3： 17 | ``` 18 | 输入：[1, 4, 6]，target = 5, 19 | 输出： 1 或 2. 20 | ``` 21 | ### 参考代码 22 | 利用二分查找的思想即可 23 | ```python 24 | class Solution: 25 | # @param {int[]} A an integer array sorted in ascending order 26 | # @param {int} target an integer 27 | # @return {int} an integer 28 | def closestNumber(self, A, target): 29 | if not A: 30 | return -1 31 | start, end = 0, len(A) - 1 32 | while start + 1 < end: 33 | mid = (start + end) / 2 34 | if A[mid] == target: 35 | return mid 36 | elif A[mid] > target: 37 | end = mid 38 | else: 39 | start = mid 40 | if A[end] - target > target - A[start]: 41 | return start 42 | else: 43 | return end 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/二分法/maximum-average-subarray.md: -------------------------------------------------------------------------------- 1 | # 子数组的最大平均值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/maximum-average-subarray/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个由`n`个整数组成的数组，找到给定长度`k`的连续子数组，该子数组具有最大平均值。你需要输出最大平均值。 5 | ### 样例说明 6 | **样例1** 7 | 8 | ``` 9 | 输入: nums = [1,12,-5,-6,50,3] and k = 4 10 | 输出: 12.75 11 | 解释: 12 | 最大平均为(12-5-6+50)/4 = 51/4 = 12.75 13 | ``` 14 | 15 | 16 | 17 | **样例2** 18 | 19 | ``` 20 | 输入: nums = [4,2,1,3,3] and k = 2 21 | 输出: 3.00 22 | 解释: 23 | 最大平均为(3+3)/2 = 6/2 = 3.00 24 | ``` 25 | ### 参考代码 26 | ```java 27 | public class Solution { 28 | /** 29 | * @param nums: an array 30 | * @param k: an integer 31 | * @return: the maximum average value 32 | */ 33 | public double findMaxAverage(int[] nums, int k) { 34 | // Write your code here 35 | int n = nums.length; 36 | int[] sum = new int[n + 1]; 37 | for(int i = 1; i <= n; i++){ 38 | sum[i] = sum[i - 1] + nums[i - 1]; 39 | } 40 | int ans = Integer.MIN_VALUE; 41 | for(int i = k; i <= n; i++){ 42 | ans = Math.max(ans, sum[i] - sum[i - k]); 43 | } 44 | return ans * 1.0 / k; 45 | } 46 | } 47 | ``` 48 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/二分法/sqrtx-ii.md: -------------------------------------------------------------------------------- 1 | # 对x开根II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/sqrtx-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 实现 `double sqrt(double x)` 并且 `x >= 0`。 5 | 计算并返回x开根后的值。 6 | ### 样例说明 7 | 例1: 8 | ``` 9 | 输入: n = 2 10 | 输出: 1.41421356 11 | ``` 12 | 例2: 13 | ``` 14 | 输入: n = 3 15 | 输出: 1.73205081 16 | ``` 17 | 18 | ### 参考代码 19 | ```python 20 | class Solution: 21 | """ 22 | @param: x: a double 23 | @return: the square root of x 24 | """ 25 | def sqrt(self, x): 26 | if x >= 1: 27 | start, end = 1, x 28 | else: 29 | start, end = x, 1 30 | 31 | while end - start > 1e-10: 32 | mid = (start + end) / 2 33 | if mid * mid < x: 34 | start = mid 35 | else: 36 | end = mid 37 | 38 | return start 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/add-strings.md: -------------------------------------------------------------------------------- 1 | # 大整数加法 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/add-strings/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 以字符串的形式给出两个非负整数 `num1` 和 `num2`，返回 `num1` 和 `num2` 的和。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入 : num1 = "123", num2 = "45" 9 | 输出 : "168" 10 | ``` 11 | ### 参考代码 12 | 将两个数解析后相加即可 13 | ```java 14 | public class Solution { 15 | public String addStrings(String num1, String num2) { 16 | String res = ""; 17 | int m = num1.length(), n = num2.length(), i = m - 1, j = n - 1, flag = 0; 18 | while(i >= 0 || j >= 0){ 19 | int a, b; 20 | if(i >= 0){ 21 | a = num1.charAt(i--) - '0'; 22 | } 23 | else{ 24 | a = 0; 25 | } 26 | if(j >= 0){ 27 | b = num2.charAt(j--) - '0'; 28 | } 29 | else{ 30 | b = 0; 31 | } 32 | int sum = a + b + flag; 33 | res =(char)(sum % 10 + '0') + res; 34 | flag = sum / 10; 35 | } 36 | return flag == 1 ? "1" + res: res; 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/bitwise-and-of-numbers-range.md: -------------------------------------------------------------------------------- 1 | # 数字范围内位的与 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/bitwise-and-of-numbers-range/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定范围[m，n]，其中0 <= m <= n <= 2147483647，输出此范围内所有数字的按位AND，包括端值。 5 | 6 | 例如，给定范围[5,7]，输出返回4。 7 | ### 样例说明 8 | **样例1** 9 | 10 | ``` 11 | 输入: m=5, n=7 12 | 输出: 4 13 | ``` 14 | 15 | **样例2** 16 | 17 | ``` 18 | 输入: m=14, n=15 19 | 输出: 14 20 | ``` 21 | ### 参考代码 22 | 本题中，我们需要得到[m,n]所有元素按位与的结果。举个例子，当m=26，n=30时，它们的二进制表示为为： 23 | **11**010　　**11**011　　**11**100　　**11**101　　**11**110 24 | 这个样例的答案是**11**000，易见我们发现我们只需要找到m和n最左边的公共部分即可。 25 | 26 | 每次都将n与n-1按位与，当n的二进制为1010时，1010 & 1001 = 1000，相当于把二进制位的最后一个1去掉了。因此我们不断的做n^n-1的操作，直到n小于m相等即可。 27 | ```java 28 | public class Solution { 29 | /** 30 | * @param m: an Integer 31 | * @param n: an Integer 32 | * @return: the bitwise AND of all numbers in [m,n] 33 | */ 34 | public int rangeBitwiseAnd(int m, int n) { 35 | // Write your code here 36 | while(m < n) { 37 | n = n & (n-1); 38 | } 39 | return n; 40 | 41 | } 42 | } 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/count-1-in-binary.md: -------------------------------------------------------------------------------- 1 | # 二进制中有多少个1 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-1-in-binary/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算在一个 32 位的整数的二进制表示中有多少个 `1`。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：32 9 | 输出：1 10 | 解释： 11 | 32(100000)，返回 1。 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：5 16 | 输出：2 17 | 解释： 18 | 5(101)，返回 2。 19 | ``` 20 | ### 参考代码 21 | ```java 22 | public class Solution { 23 | /** 24 | * @param num: an integer 25 | * @return: an integer, the number of ones in num 26 | */ 27 | public int countOnes(int num) { 28 | int count = 0; 29 | while (num != 0) { 30 | num = num & (num - 1); 31 | count++; 32 | } 33 | return count; 34 | } 35 | }; 36 | 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/flip-bits.md: -------------------------------------------------------------------------------- 1 | # 将整数A转换为B 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/flip-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 如果要将整数n转换为m，需要改变多少个bit位？ 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | a ^ b 得到的结果的二进制中，1的个数就是a和b相异的位数。 9 | ```java 10 | class Solution { 11 | /** 12 | *@param a, b: Two integer 13 | *return: An integer 14 | */ 15 | public static int bitSwapRequired(int a, int b) { 16 | // write your code here 17 | int count = 0; 18 | for (int c = a ^ b; c != 0; c = c >>> 1) { 19 | count += c & 1; 20 | } 21 | return count; 22 | } 23 | }; 24 | ``` 25 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/hamming-distance.md: -------------------------------------------------------------------------------- 1 | # Hamming距离 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/hamming-distance/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 两个整数的Hamming距离是对应比特位不同的个数。 5 | 给定两个整数x和y，计算两者的Hamming距离。 6 | ### 样例说明 7 | **样例1** 8 | 9 | ``` 10 | 输入: x = 1 和 y = 4 11 | 输出: 2 12 | 解释: 13 | 1的二进制表示是001 14 | 4的二进制表示是100 15 | 共有2位不同 16 | ``` 17 | 18 | 19 | 20 | **样例2** 21 | 22 | ``` 23 | 输入: x = 5 和 y = 2 24 | 输出: 3 25 | 解释: 26 | 5的二进制表示是101 27 | 2的二进制表示是010 28 | 共有3位不同 29 | ``` 30 | ### 参考代码 31 | ```java 32 | public class Solution { 33 | public int hammingDistance(int x, int y) { 34 | int Distance=0; 35 | 36 | while ( x != 0 || y != 0 ) { 37 | if ( x % 2 != y % 2 ) { 38 | Distance ++; 39 | } 40 | x /= 2; 41 | y /= 2; 42 | } 43 | return Distance; 44 | } 45 | } 46 | ``` 47 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/number-of-1-bits.md: -------------------------------------------------------------------------------- 1 | # 判断一个整数中有多少个1 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/number-of-1-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 写一个函数，其以无符号整数为输入，而输出对应二进制数所具有的“1”的位数（也被称为汉明权重） 5 | ### 样例说明 6 | **样例 1** 7 | ``` 8 | 输入：n = 11 9 | 输出：3 10 | 解析：11(10) = 1011(2), 返回 3 11 | ``` 12 | **样例 2** 13 | ``` 14 | 输入：n = 7 15 | 输出：3 16 | 解析：7(10) = 111(2), 返回 3 17 | ``` 18 | ### 参考代码 19 | 使用位运算符号‘&’，与1做与操作 20 | 并将n右移一位( >>1 ) 21 | 直到n为0 22 | ```java 23 | public class Solution { 24 | /** 25 | * @param n: an unsigned integer 26 | * @return: the number of ’1' bits 27 | */ 28 | public int hammingWeight(int n) { 29 | int ones = 0; 30 | while(n!=0) { 31 | ones += (n & 1); 32 | n = n>>1; 33 | } 34 | return ones; 35 | } 36 | } 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/o1-check-power-of-2.md: -------------------------------------------------------------------------------- 1 | # O(1)时间检测2的幂次 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/o1-check-power-of-2/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用 O(*1*) 时间检测整数 *n* 是否是 *2* 的幂次。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | 2的幂的数字，其二进制中只有一个1。 9 | $n&(n-1) == 0$ 说明只有1个1。 10 | ```cpp 11 | class Solution { 12 | public: 13 | /* 14 | * @param n: An integer 15 | * @return: True or false 16 | */ 17 | bool checkPowerOf2(int n) { 18 | // write your code here 19 | if (n==0) 20 | return false; 21 | while (n % 2 == 0) 22 | n /=2 ; 23 | return n == 1; 24 | } 25 | }; 26 | ``` 27 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/single-number.md: -------------------------------------------------------------------------------- 1 | # 落单的数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/single-number/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出 `2 * n + 1`个数字，除其中一个数字之外其他每个数字均出现两次，找到这个数字。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入：[1,1,2,2,3,4,4] 9 | 输出：3 10 | 解释： 11 | 仅3出现一次 12 | ``` 13 | **样例 2:** 14 | ``` 15 | 输入：[0,0,1] 16 | 输出：1 17 | 解释： 18 | 仅1出现一次 19 | ``` 20 | 21 | ### 参考代码 22 | Given an array of integers, every element appears twice except for one. Find that single one. 23 | 考点： 24 | * 异或位运算 25 | 26 | Note: 27 | Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 28 | 异或运算具有很好的性质，相同数字异或运算后为0，并且具有交换律和结合律，故将所有数字异或运算后即可得到只出现一次的数字。 29 | ```python 30 | class Solution: 31 | """ 32 | @param A : an integer array 33 | @return : a integer 34 | """ 35 | def singleNumber(self, A): 36 | # write your code here 37 | ans = 0; 38 | for x in A: 39 | ans = ans ^ x 40 | return ans 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/swap-bits.md: -------------------------------------------------------------------------------- 1 | # 交换奇偶二进制位 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/swap-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 设计一个方法，用尽可能少的指令，将一个整数中奇数数位和偶数数位的数字交换 （如，数位 0 和数位 1 交换，数位 2 和数位 3 交换，等等）。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：0 9 | 输出：0 10 | 解释： 11 | 0 = 0(2) -> 0(2) = 0 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：5 16 | 输出：10 17 | 解释： 18 | 5 = 101(2) -> 1010(2) = 10 19 | ``` 20 | ### 参考代码 21 | ```cpp 22 | class Solution { 23 | public: 24 | /** 25 | * @param x an integer 26 | * @return an integer 27 | */ 28 | int swapOddEvenBits(int x) { 29 | // Write your code here 30 | return ( ((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1) ); 31 | } 32 | }; 33 | ``` 34 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/位运算/swap-without-extra-variable-only-c.md: -------------------------------------------------------------------------------- 1 | # 交换不借助额外变量(仅C++) 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/swap-without-extra-variable-only-c/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出两个变量，x 和 y ，不使用第三个变量交换这两个变量的值。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入 : x = 10, y = 5 9 | 输出 : x = 5, y = 10 10 | ``` 11 | ### 参考代码 12 | 通过位运算实现。 13 | ```cpp 14 | class Solution { 15 | public: 16 | /** 17 | * @param x an integer 18 | * @param y an integer 19 | * @return nothing 20 | */ 21 | void swap(int &x, int &y) { 22 | // Write your code here 23 | x = x ^ y; 24 | y = x ^ y; 25 | x = x ^ y; 26 | } 27 | }; 28 | ``` 29 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/max-of-3-numbers.md: -------------------------------------------------------------------------------- 1 | # 三数之中的最大值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/max-of-3-numbers/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给三个整数，求他们中的最大值。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: num1 = 1, num2 = 9, num3 = 0 9 | 输出: 9 10 | 11 | 样例解释: 12 | 返回三个数中最大的数。 13 | 14 | 样例 2: 15 | 输入: num1 = 1, num2 = 2, num3 = 3 16 | 输出: 3 17 | 18 | 样例解释: 19 | 返回三个中最大的数字。 20 | 21 | ``` 22 | ### 参考代码 23 | if选择语句的基本操作 24 | ```java 25 | public class Solution { 26 | /** 27 | * @param a an integer 28 | * @param b an integer 29 | * @param c an integer 30 | * @return an integer 31 | */ 32 | public int maxOfThreeNumbers(int a, int b, int c) { 33 | if (a >= b && a >= c) { 34 | return a; 35 | } else if (b >= a && b >= c) { 36 | return b; 37 | } else { 38 | return c; 39 | } 40 | } 41 | } 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/remove-element.md: -------------------------------------------------------------------------------- 1 | # 删除元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/remove-element/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个数组和一个值，在原地删除与值相同的数字，返回新数组的长度。

元素的顺序可以改变，并且对新的数组不会有影响。

5 | ### 样例说明 6 | 7 | ### 参考代码 8 | Given an array and a value, remove all instances of that value in place and return the new length. 9 | 10 | The order of elements can be changed. It doesn't matter what you leave beyond the new length. 11 | 12 | 从前往后，遇到等于value的就跳过，把后面的元素填到前面的位置里。 13 | ```java 14 | public class Solution { 15 | public int removeElement(int[] A, int elem) { 16 | int i = 0; 17 | int pointer = A.length - 1; 18 | while(i <= pointer){ 19 | if(A[i] == elem){ 20 | A[i] = A[pointer]; 21 | pointer--; 22 | } else { 23 | i++; 24 | } 25 | } 26 | return pointer + 1; 27 | } 28 | } 29 | ``` 30 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/reverse-3-digit-integer.md: -------------------------------------------------------------------------------- 1 | # 反转一个3位整数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reverse-3-digit-integer/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 反转一个只有3位数的整数。 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: number = 123 10 | 输出: 321 11 | ``` 12 | **样例 2:** 13 | 14 | ``` 15 | 输入: number = 900 16 | 输出: 9 17 | ``` 18 | 19 | ### 参考代码 20 | 硅谷求职算法集训营上课版本 21 | ```java 22 | public class Solution { 23 | /* 24 | * @param number: A 3-digit number. 25 | * @return: Reversed number. 26 | */ 27 | public int reverseInteger(int number) { 28 | // write your code here 29 | //获得个位数 30 | int num1 = number % 10; 31 | //获得十位数 32 | int num2 = (number / 10) % 10; 33 | //获得百位数 34 | int num3 = ((number / 10) / 10) % 10; 35 | //相加 36 | return num3 + num2 * 10 + num1 * 100; 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/reverse-bits.md: -------------------------------------------------------------------------------- 1 | # 颠倒二进制位 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reverse-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 颠倒给定的 32 位无符号整数的二进制位。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 00000010100101000001111010011100 9 | 输出: 00111001011110000010100101000000 10 | 解释: 11 | 输入的二进制串 00000010100101000001111010011100 表示无符号整数 43261596， 12 | 因此返回 964176192，其二进制表示形式为 00111001011110000010100101000000。 13 | ``` 14 | **样例 2:** 15 | ``` 16 | 输入：11111111111111111111111111111101 17 | 输出：10111111111111111111111111111111 18 | 解释： 19 | 输入的二进制串 11111111111111111111111111111101 表示无符号整数 4294967293， 20 | 因此返回 3221225471 其二进制表示形式为 10101111110010110010011101101001。 21 | ``` 22 | ### 参考代码 23 | 分离出输入数字的每一位，最低位左移31bit，次低位左移30bit，以此类推。 24 | ```java 25 | public class Solution { 26 | // you need treat n as an unsigned value 27 | public long reverseBits(int n) { 28 | int reversed = 0; 29 | for (int i = 0; i < 32; i++) { 30 | reversed = (reversed << 1) | (n & 1); 31 | n = (n >> 1); 32 | } 33 | return reversed; 34 | } 35 | } 36 | ``` 37 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/search-in-rotated-sorted-array-ii.md: -------------------------------------------------------------------------------- 1 | # 搜索旋转排序数组 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/search-in-rotated-sorted-array-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

跟进“搜索旋转排序数组”，假如有重复元素又将如何？

是否会影响运行时间复杂度？

如何影响？

为何会影响？

写出一个函数判断给定的目标值是否出现在数组中。

5 | ### 样例说明 6 | 例1: 7 | ``` 8 | 输入: 9 | [] 10 | 1 11 | 输出: 12 | false 13 | ``` 14 | 例2: 15 | ``` 16 | 输入: 17 | [3,4,4,5,7,0,1,2] 18 | 4 19 | 输出: 20 | true 21 | ``` 22 | 23 | ### 参考代码 24 | 由于出现重复元素，故不能使用二分法。 25 | 所以暴力比较即可。 26 | ```java 27 | public class Solution { 28 | // 这个问题在面试中不会让实现完整程序 29 | // 只需要举出能够最坏情况的数据是 [1,1,1,1... 1] 里有一个0即可。 30 | // 在这种情况下是无法使用二分法的，复杂度是O(n) 31 | // 因此写个for循环最坏也是O(n)，那就写个for循环就好了 32 | // 如果你觉得，不是每个情况都是最坏情况，你想用二分法解决不是最坏情况的情况，那你就写一个二分吧。 33 | // 反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。 34 | public boolean search(int[] A, int target) { 35 | for (int i = 0; i < A.length; i ++) { 36 | if (A[i] == target) { 37 | return true; 38 | } 39 | } 40 | return false; 41 | } 42 | } 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/singleton.md: -------------------------------------------------------------------------------- 1 | # 单例 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/singleton/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | **单例** 是最为最常见的设计模式之一。对于任何时刻，如果某个类只存在且最多存在一个具体的实例，那么我们称这种设计模式为单例。例如，对于 class Mouse (不是动物的mouse哦)，我们应将其设计为 singleton 模式。 5 | 6 | 你的任务是设计一个 `getInstance` 方法，对于给定的类，每次调用 `getInstance` 时，都可得到同一个实例。 7 | ### 样例说明 8 | ``` 9 | 在 Java 中: 10 | 11 | A a = A.getInstance(); 12 | A b = A.getInstance(); 13 | 14 | a 应等于 b. 15 | 16 | ``` 17 | ### 参考代码 18 | 创建私有静态成员对象，每次获取时返回此对象。 19 | ```java 20 | class Solution { 21 | 22 | 23 | /** 24 | * @return: The same instance of this class every time 25 | */ 26 | public static Solution instance = null; 27 | public static Solution getInstance() { 28 | if (instance == null) { 29 | instance = new Solution(); 30 | } 31 | return instance; 32 | } 33 | }; 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/trailing-zeros.md: -------------------------------------------------------------------------------- 1 | # 尾部的零 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/trailing-zeros/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

设计一个算法，计算出n阶乘中尾部零的个数

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 11 9 | 输出: 2 10 | 11 | 样例解释: 12 | 11! = 39916800, 结尾的0有2个。 13 | 14 | 样例 2: 15 | 输入: 5 16 | 输出: 1 17 | 18 | 样例解释: 19 | 5! = 120， 结尾的0有1个。 20 | 21 | ``` 22 | ### 参考代码 23 | 可以将每个数拆分成其素因子的乘积，可以发现，0是由`2*5`产生的，而5的数量一定小于2的数量，因此5的个数决定了结尾0的个数。 24 | 25 | 只要计算n的阶乘中，5这个素因子出现多少次即可。 26 | ```java 27 | class Solution { 28 | /* 29 | * param n: As desciption 30 | * return: An integer, denote the number of trailing zeros in n! 31 | */ 32 | public long trailingZeros(long n) { 33 | long sum = 0; 34 | while (n != 0) { 35 | sum += n / 5; 36 | n /= 5; 37 | } 38 | return sum; 39 | } 40 | }; 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/其他/ugly-number.md: -------------------------------------------------------------------------------- 1 | # 丑数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/ugly-number/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 写一个程序来检测一个整数是不是`丑数`。 5 | 6 | 丑数的定义是，只包含质因子 `2, 3, 5` 的正整数。比如 6, 8 就是丑数，但是 14 不是丑数因为他包含了质因子 7。 7 | ### 样例说明 8 | 例1: 9 | ``` 10 | 输入: num = 8 11 | 输出: true 12 | 解释: 13 | 8=2*2*2 14 | ``` 15 | 例2: 16 | ``` 17 | 输入: num = 14 18 | 输出: false 19 | 解释: 20 | 14=2*7 21 | ``` 22 | 23 | ### 参考代码 24 | 看这个数是否只有2，3，5的因子 25 | ```java 26 | public class Solution { 27 | /** 28 | * @param num an integer 29 | * @return true if num is an ugly number or false 30 | */ 31 | public boolean isUgly(int num) { 32 | if (num <= 0) return false; 33 | if (num == 1) return true; 34 | 35 | while (num >= 2 && num % 2 == 0) num /= 2; 36 | while (num >= 3 && num % 3 == 0) num /= 3; 37 | while (num >= 5 && num % 5 == 0) num /= 5; 38 | 39 | return num == 1; 40 | } 41 | } 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/双指针/linked-list-cycle.md: -------------------------------------------------------------------------------- 1 | # 带环链表 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/linked-list-cycle/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个链表，判断它是否有环。

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 21->10->4->5, then tail connects to node index 1(value 10). 9 | 输出: true 10 | 11 | 样例 2: 12 | 输入: 21->10->4->5->null 13 | 输出: false 14 | 15 | ``` 16 | 17 | ### 参考代码 18 | Given a linked list, determine if it has a cycle in it. 19 | 20 | Follow up: 21 | Can you solve it without using extra space? 22 | 23 | 快慢指针的经典题。 24 | 快指针每次走两步，慢指针一次走一步。 25 | 在慢指针进入环之后，快慢指针之间的距离每次缩小1，所以最终能相遇。 26 | ```java 27 | public class Solution { 28 | public Boolean hasCycle(ListNode head) { 29 | if (head == null || head.next == null) { 30 | return false; 31 | } 32 | 33 | ListNode fast, slow; 34 | fast = head.next; 35 | slow = head; 36 | while (fast != slow) { 37 | if(fast==null || fast.next==null) 38 | return false; 39 | fast = fast.next.next; 40 | slow = slow.next; 41 | } 42 | return true; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/双指针/window-sum.md: -------------------------------------------------------------------------------- 1 | # 滑动窗口内数的和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/window-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你一个大小为n的整型数组和一个大小为k的滑动窗口，将滑动窗口从头移到尾，输出从开始到结束每一个时刻滑动窗口内的数的和。 5 | ### 样例说明 6 | **样例 1** 7 | ``` 8 | 输入：array = [1,2,7,8,5], k = 3 9 | 输出：[10,17,20] 10 | 解析： 11 | 1 + 2 + 7 = 10 12 | 2 + 7 + 8 = 17 13 | 7 + 8 + 5 = 20 14 | ``` 15 | ### 参考代码 16 | ```python 17 | class Solution: 18 | # @param nums {int[]} a list of integers 19 | # @param k {int} size of window 20 | # @return {int[]} the sum of element inside the window at each moving 21 | def winSum(self, nums, k): 22 | # Write your code here 23 | n = len(nums) 24 | if n < k or k <= 0: 25 | return [] 26 | sums = [0] * (n - k + 1) 27 | for i in range(k): 28 | sums[0] += nums[i]; 29 | 30 | for i in range(1, n - k + 1): 31 | sums[i] = sums[i - 1] - nums[i - 1] + nums[i + k - 1] 32 | 33 | return sums 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/字典MAP/string-permutation.md: -------------------------------------------------------------------------------- 1 | # 字符串置换 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/string-permutation/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定两个字符串，请设计一个方法来判定其中一个字符串是否为另一个字符串的置换。 5 | 6 | 置换的意思是，通过改变顺序可以使得两个字符串相等。 7 | ### 样例说明 8 | 9 | ### 参考代码 10 | 统计字符数量，组成成分一致即可置换得到。 11 | ```java 12 | public class Solution { 13 | /** 14 | * @param A a string 15 | * @param B a string 16 | * @return a boolean 17 | */ 18 | public boolean stringPermutation(String A, String B) { 19 | // Write your code here 20 | int[] cnt = new int[1000]; 21 | for (int i = 0; i < A.length(); ++i) 22 | cnt[(int)A.charAt(i)] += 1; 23 | for (int i = 0; i < B.length(); ++i) 24 | cnt[(int)B.charAt(i)] -= 1; 25 | for (int i = 0; i < 1000; ++i) 26 | if (cnt[i] != 0) 27 | return false; 28 | return true; 29 | } 30 | } 31 | ``` 32 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/字典MAP/unique-characters.md: -------------------------------------------------------------------------------- 1 | # 判断字符串是否没有重复字符 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/unique-characters/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 实现一个算法确定字符串中的字符是否均唯一出现 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: "abc_____" 10 | 输出: false 11 | ``` 12 | 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: "abc" 17 | 输出: true 18 | ``` 19 | ### 参考代码 20 | 使用一个map或数组标记每个字符是否出现过。 21 | ```java 22 | public class Solution { 23 | /** 24 | * @param str: a string 25 | * @return: a boolean 26 | */ 27 | public boolean isUnique(String str) { 28 | // write your code here 29 | boolean[] char_set = new boolean[256]; 30 | for (int i = 0; i < str.length(); i++) { 31 | int val = str.charAt(i); 32 | if (char_set[val]) return false; 33 | char_set[val] = true; 34 | } 35 | return true; 36 | } 37 | } 38 | ``` 39 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/排序算法/insertion-sort-list.md: -------------------------------------------------------------------------------- 1 | # 链表插入排序 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/insertion-sort-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用插入排序对链表排序 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 0->null 9 | 输出: 0->null 10 | 11 | 12 | 样例 2: 13 | 输入: 1->3->2->0->null 14 | 输出 :0->1->2->3->null 15 | 16 | 17 | ``` 18 | 19 | ### 参考代码 20 | Sort a linked list using insertion sort. 21 | 22 | 与数组普通插入排序一样的做法。 23 | ```java 24 | public class Solution { 25 | public ListNode insertionSortList(ListNode head) { 26 | ListNode dummy = new ListNode(0); 27 | // 这个dummy的作用是，把head开头的链表一个个的插入到dummy开头的链表里 28 | // 所以这里不需要dummy.next = head; 29 | 30 | while (head != null) { 31 | ListNode node = dummy; 32 | while (node.next != null && node.next.val < head.val) { 33 | node = node.next; 34 | } 35 | ListNode temp = head.next; 36 | head.next = node.next; 37 | node.next = head; 38 | head = temp; 39 | } 40 | 41 | return dummy.next; 42 | } 43 | } 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/排序算法/recover-rotated-sorted-array.md: -------------------------------------------------------------------------------- 1 | # 恢复旋转排序数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/recover-rotated-sorted-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个**旋转**排序数组，在原地恢复其排序。（升序） 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | Given a rotated sorted array, recover it to sorted array in-place. 9 | 10 | Example 11 | [4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5] 12 | ```python 13 | class Solution: 14 | """ 15 | @param nums: The rotated sorted array 16 | @return: The recovered sorted array 17 | """ 18 | def recoverRotatedSortedArray(self, nums): 19 | nums.sort() 20 | ``` 21 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/排序算法/remove-duplicates-from-sorted-list.md: -------------------------------------------------------------------------------- 1 | # 删除排序链表中的重复元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/remove-duplicates-from-sorted-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个排序链表，删除所有重复的元素每个元素只留下一个。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: null 9 | 输出: null 10 | 11 | 12 | 样例 2: 13 | 输入: 1->1->2->null 14 | 输出: 1->2->null 15 | 16 | 样例 3: 17 | 输入: 1->1->2->3->3->null 18 | 输出: 1->2->3->null 19 | 20 | 21 | ``` 22 | ### 参考代码 23 | Given a sorted linked list, delete all duplicates such that each element appear only once. 24 | 25 | For example, 26 | Given 1->1->2, return 1->2. 27 | Given 1->1->2->3->3, return 1->2->3. 28 | 29 | 跳过相同的元素，指向再后一个元素。 30 | ```java 31 | public class Solution { 32 | public ListNode deleteDuplicates(ListNode head) { 33 | if (head == null) { 34 | return null; 35 | } 36 | 37 | ListNode node = head; 38 | while (node.next != null) { 39 | if (node.val == node.next.val) { 40 | node.next = node.next.next; 41 | } else { 42 | node = node.next; 43 | } 44 | } 45 | return head; 46 | } 47 | } 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/排序算法/sort-letters-by-case.md: -------------------------------------------------------------------------------- 1 | # 字符大小写排序 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/sort-letters-by-case/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个只包含字母的字符串，按照先小写字母后大写字母的顺序进行排序。

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: "abAcD" 9 | 输出: "acbAD" 10 | 11 | 样例 2: 12 | 输入: "ABC" 13 | 输出: "ABC" 14 | 15 | ``` 16 | ### 参考代码 17 | ```java 18 | public class Solution { 19 | public void sortLetters(char[] chars) { 20 | int i = 0, j = chars.length - 1; 21 | 22 | while (i <= j) { 23 | while (i <= j && Character.isLowerCase(chars[i])) { 24 | i++; 25 | } 26 | while (i <= j && Character.isUpperCase(chars[j])) { 27 | j--; 28 | } 29 | if (i <= j) { 30 | char tmp = chars[i]; 31 | chars[i] = chars[j]; 32 | chars[j] = tmp; 33 | i++; 34 | j--; 35 | } 36 | } 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/贪心算法/assign-cookies.md: -------------------------------------------------------------------------------- 1 | # 分饼干 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/assign-cookies/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 假设你是一个了不起的家长，准备给你的孩子们一些饼干吃，但是你只能给每个孩子至多一块饼干。每一块饼干$j$都有一个尺寸$s_{j}$；同时每一个孩子$i$都有一个贪吃指数$g_{i}$，代表了能使他满足的最小的饼干尺寸。如果$s_{j} \geq g_{i}$，那么就可以将饼干$j$分给孩子$i$使他得到满足。你的目标是使最多的孩子得到满足，输出这个最大值。 5 | ### 样例说明 6 | **样例1：** 7 | 8 | ``` 9 | 输入：[1,2,3], [1,1] 10 | 输出：1 11 | 说明：你有三个孩子和两块饼干，三个孩子的贪吃指数分别是1，2，3 12 | 虽然你有两块饼干，但是因为它们的大小都为1，你只能满足让贪吃指数为1的孩子满足，因此你应该输出1 13 | ``` 14 | 15 | **样例2：** 16 | 17 | ``` 18 | 输入：[1,2], [1,2,3] 19 | 输出：2 20 | 说明：你有两个孩子和三块饼干，两个孩子的贪吃指数分别是1和2 21 | 这三块饼干的大小足以满足所有的孩子，因此你应该输出2 22 | ``` 23 | ### 参考代码 24 |


25 | ```cpp 26 | class Solution { 27 | public: 28 | int findContentChildren(vector& g, vector& s) { 29 | sort(s.begin(), s.end()); 30 | sort(g.begin(), g.end()); 31 | int i, j; 32 | for (i = j = 0; i < g.size() && j < s.size(); j++) { 33 | if (g[i] <= s[j]) { 34 | i++; 35 | } 36 | } 37 | return i; 38 | } 39 | }; 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法思维题/贪心算法/course-schedule-iii.md: -------------------------------------------------------------------------------- 1 | # 课程表 III 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/course-schedule-iii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 这里有 `n` 门不同的线上课程, 编号为 `1` 到 `n`. 每一节课都有持续时间(课程长度) `t` 和 在第 `d` 天关闭. 课程应连续持续 `t` 天，必须在第`d `天或之前完成, 你将从第一天开始. 5 | 给出 `n` 门线上课程用 `pairs (t, d)` 来表示, 你的任务是找到可以上的最大数量的课程数. 6 | ### 样例说明 7 | 8 | ### 参考代码 9 | 贪心法 Greedy 10 | 课程按照 deadline 排序，从左到右扫描每个课程，依次学习。如果发现学了之后超过 deadline 的，就从之前学过的课程里扔掉一个耗时最长的。 11 | 因为这样可以使得其他的课程往前挪，而往前挪是没影响的。 12 | 所以挑最大值这个事情就是 Heap 的事情了 13 | ```python 14 | import heapq 15 | 16 | 17 | class Solution: 18 | """ 19 | @param courses: duration and close day of each course 20 | @return: the maximal number of courses that can be taken 21 | """ 22 | def scheduleCourse(self, courses): 23 | courses = sorted(courses, key=lambda x: x[1]) 24 | curt_time = 0 25 | heap = [] 26 | for duration, deadline in courses: 27 | curt_time += duration 28 | heapq.heappush(heap, -duration) 29 | if curt_time > deadline: 30 | curt_time -= (-heapq.heappop(heap)) 31 | 32 | return len(heap) 33 | ``` 34 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /算法面试题解/Google 面试题 | 找二叉树最底层最左边的节点.md: -------------------------------------------------------------------------------- 1 | ## 题目描述 2 | 3 | 给一棵二叉树，求出该二叉树最底层最左边的节点。 4 | 5 | **样例:** 6 | 7 | ![image](https://github.com/ninechapter-algorithm/ninechapter-algorithm/blob/master/pictures/%E6%89%BE%E4%BA%8C%E5%8F%89%E6%A0%91%E6%9C%80%E5%BA%95%E5%B1%82%E6%9C%80%E5%B7%A6%E8%BE%B9%E7%9A%84%E8%8A%82%E7%82%B9-%E6%A0%B7%E4%BE%8B.png) 8 | 9 | ## 解题思路分析 10 | 11 | 这题有2个做法，一是宽度优先搜索，用每层的第一个节点来更新答案。二是深度优先搜索，当遇到一个节点的深度大于目前维护的最大深度时用这个节点来更新答案。 12 | 13 | 1. 使用**宽度优先搜索**bfs，用每层的第一个节点更新Ans。时间复杂度O(n)。 14 | 15 | 2. 使用**深度优先搜索**dfs，当我们第一次访问一个深度为depth的节点x（之前只访问过深度小于depth的节点）时，x一定是depth深度的最左节点，用这个节点更新Ans。即我们维护一个最大深度，当遍历到一个点的深度大于最大深度时，用这个节点来更新答案，并更新最大深度即可。时间复杂度O(n)。 16 | 17 | ## 参考程序 18 | 19 | ![image](https://github.com/ninechapter-algorithm/ninechapter-algorithm/blob/master/pictures/%E6%89%BE%E4%BA%8C%E5%8F%89%E6%A0%91%E6%9C%80%E5%BA%95%E5%B1%82%E6%9C%80%E5%B7%A6%E8%BE%B9%E7%9A%84%E8%8A%82%E7%82%B9.%E9%A2%98%E8%A7%A3.png) 20 | 21 | 更多题解参考：[LintCode/LeetCode 参考答案查询](http://www.jiuzhang.com/solution/find-bottom-left-tree-value/?utm_source=sc-github-lm) 22 | 23 | ## 面试官角度分析 24 | 25 | 这道题是一道**简单难度题**，考点是基础数据结构二叉树和在二叉树中的搜索，用bfs和dfs均可。给出正确算法即可达到hire。 26 | 27 | ## LintCode相关问题 28 | 29 | [69. 二叉树的层次遍历](https://www.lintcode.com/problem/binary-tree-level-order-traversal/?utm_source=sc-github-lm) 30 | 31 | -------------------------------------------------------------------------------- /编程新手必刷50题 V1.0.docx: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/huangyingw/ninechapter-algorithm_linghu-algorithm-templete/ff7284350de9097925a41b59f293f1f1dda25622/编程新手必刷50题 V1.0.docx -------------------------------------------------------------------------------- /面试常考算法必刷题库.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/huangyingw/ninechapter-algorithm_linghu-algorithm-templete/ff7284350de9097925a41b59f293f1f1dda25622/面试常考算法必刷题库.pdf -------------------------------------------------------------------------------- /题解/add-strings.md: -------------------------------------------------------------------------------- 1 | # 大整数加法 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/add-strings/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 以字符串的形式给出两个非负整数 `num1` 和 `num2`，返回 `num1` 和 `num2` 的和。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入 : num1 = "123", num2 = "45" 9 | 输出 : "168" 10 | ``` 11 | ### 参考代码 12 | 将两个数解析后相加即可 13 | ```java 14 | public class Solution { 15 | public String addStrings(String num1, String num2) { 16 | String res = ""; 17 | int m = num1.length(), n = num2.length(), i = m - 1, j = n - 1, flag = 0; 18 | while(i >= 0 || j >= 0){ 19 | int a, b; 20 | if(i >= 0){ 21 | a = num1.charAt(i--) - '0'; 22 | } 23 | else{ 24 | a = 0; 25 | } 26 | if(j >= 0){ 27 | b = num2.charAt(j--) - '0'; 28 | } 29 | else{ 30 | b = 0; 31 | } 32 | int sum = a + b + flag; 33 | res =(char)(sum % 10 + '0') + res; 34 | flag = sum / 10; 35 | } 36 | return flag == 1 ? "1" + res: res; 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/anagrams.md: -------------------------------------------------------------------------------- 1 | # 乱序字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/anagrams/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给出一个字符串数组S，找到其中所有的乱序字符串(Anagram)。如果一个字符串是乱序字符串，那么他存在一个字母集合相同，但顺序不同的字符串也在S中。

5 | ### 样例说明 6 | **样例1:** 7 | ``` 8 | 输入:["lint", "intl", "inlt", "code"] 9 | 输出:["lint", "inlt", "intl"] 10 | ``` 11 | 12 | **样例 2:** 13 | ``` 14 | 输入:["ab", "ba", "cd", "dc", "e"] 15 | 输出: ["ab", "ba", "cd", "dc"] 16 | ``` 17 | ### 参考代码 18 | Given an array of strings, return all groups of strings that are anagrams. 19 | 20 | Note: All inputs will be in lower-case 21 | ```python 22 | class Solution: 23 | # @param strs: A list of strings 24 | # @return: A list of strings 25 | def anagrams(self, strs): 26 | # write your code here 27 | dict = {} 28 | for word in strs: 29 | sortedword = ''.join(sorted(word)) 30 | dict[sortedword] = [word] if sortedword not in dict else dict[sortedword] + [word] 31 | res = [] 32 | for item in dict: 33 | if len(dict[item]) >= 2: 34 | res += dict[item] 35 | return res 36 | 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/assign-cookies.md: -------------------------------------------------------------------------------- 1 | # 分饼干 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/assign-cookies/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 假设你是一个了不起的家长，准备给你的孩子们一些饼干吃，但是你只能给每个孩子至多一块饼干。每一块饼干$j$都有一个尺寸$s_{j}$；同时每一个孩子$i$都有一个贪吃指数$g_{i}$，代表了能使他满足的最小的饼干尺寸。如果$s_{j} \geq g_{i}$，那么就可以将饼干$j$分给孩子$i$使他得到满足。你的目标是使最多的孩子得到满足，输出这个最大值。 5 | ### 样例说明 6 | **样例1：** 7 | 8 | ``` 9 | 输入：[1,2,3], [1,1] 10 | 输出：1 11 | 说明：你有三个孩子和两块饼干，三个孩子的贪吃指数分别是1，2，3 12 | 虽然你有两块饼干，但是因为它们的大小都为1，你只能满足让贪吃指数为1的孩子满足，因此你应该输出1 13 | ``` 14 | 15 | **样例2：** 16 | 17 | ``` 18 | 输入：[1,2], [1,2,3] 19 | 输出：2 20 | 说明：你有两个孩子和三块饼干，两个孩子的贪吃指数分别是1和2 21 | 这三块饼干的大小足以满足所有的孩子，因此你应该输出2 22 | ``` 23 | ### 参考代码 24 |


25 | ```cpp 26 | class Solution { 27 | public: 28 | int findContentChildren(vector& g, vector& s) { 29 | sort(s.begin(), s.end()); 30 | sort(g.begin(), g.end()); 31 | int i, j; 32 | for (i = j = 0; i < g.size() && j < s.size(); j++) { 33 | if (g[i] <= s[j]) { 34 | i++; 35 | } 36 | } 37 | return i; 38 | } 39 | }; 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/backpack-iv.md: -------------------------------------------------------------------------------- 1 | # 背包问题 IV 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack-iv/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出 n 个物品, 以及一个数组, `nums[i]`代表第i个物品的大小, 保证大小均为正数并且没有重复, 正整数 `target` 表示背包的大小, 找到能填满背包的方案数。 5 | `每一个物品可以使用无数次` 6 | ### 样例说明 7 | **样例1** 8 | 9 | ``` 10 | 输入: nums = [2,3,6,7] 和 target = 7 11 | 输出: 2 12 | 解释: 13 | 方案有: 14 | [7] 15 | [2, 2, 3] 16 | ``` 17 | 18 | **样例2** 19 | 20 | ``` 21 | 输入: nums = [2,3,4,5] 和 target = 7 22 | 输出: 3 23 | 解释: 24 | 方案有: 25 | [2, 5] 26 | [3, 4] 27 | [2, 2, 3] 28 | ``` 29 | ### 参考代码 30 | 效率比较高的版本。时间复杂度 $O(n * target)$ 31 | ```python 32 | class Solution: 33 | """ 34 | @param nums: an integer array and all positive numbers, no duplicates 35 | @param target: An integer 36 | @return: An integer 37 | """ 38 | def backPackIV(self, nums, target): 39 | dp = [0] * (target + 1) 40 | dp[0] = 1 41 | 42 | for num in nums: 43 | for i in range(num, target + 1): 44 | dp[i] += dp[i - num] 45 | 46 | return dp[target] 47 | ``` 48 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/backpack-v.md: -------------------------------------------------------------------------------- 1 | # 背包问题 V 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack-v/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出 n 个物品, 以及一个数组, `nums[i]` 代表第i个物品的大小, 保证大小均为正数, 正整数 `target` 表示背包的大小, 找到能填满背包的方案数。 5 | `每一个物品只能使用一次` 6 | ### 样例说明 7 | 给出候选物品集合 `[1,2,3,3,7]` 以及 target `7` 8 | ``` 9 | 结果的集合为: 10 | [7] 11 | [1,3,3] 12 | ``` 13 | 返回 `2` 14 | ### 参考代码 15 | 带一点优化的动态规划。 16 | 17 | - 滚动数组。优化了空间。 18 | - 前缀和。优化了时间。没有必要每次都算到 target 那么多的值。 19 | ```python 20 | class Solution: 21 | """ 22 | @param nums: an integer array and all positive numbers 23 | @param target: An integer 24 | @return: An integer 25 | """ 26 | def backPackV(self, nums, target): 27 | f = [[0] * (target + 1), [0] * (target + 1)] 28 | f[0][0] = 1 29 | n = len(nums) 30 | prefix_sum = 0 31 | for i in range(1, n + 1): 32 | f[i % 2][0] = 1 33 | prefix_sum += nums[i - 1] 34 | for j in range(1, min(target, prefix_sum) + 1): 35 | f[i % 2][j] = f[(i - 1) % 2][j] 36 | if j >= nums[i - 1]: 37 | f[i % 2][j] += f[(i - 1) % 2][j - nums[i - 1]] 38 | 39 | return f[n % 2][target] 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/backpack-vi.md: -------------------------------------------------------------------------------- 1 | # 组合总和 IV 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack-vi/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一个都是正整数的数组 `nums`，其中没有重复的数。从中找出所有的和为 `target` 的组合个数。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | dp求解，dp[i]表示当前和为i的解法个数 9 | ```python 10 | class Solution: 11 | # @param {int[]} nums an integer array and all positive numbers, no duplicates 12 | # @param {int} target an integer 13 | # @return {int} an integer 14 | def backPackVI(self, nums, target): 15 | # Write your code here 16 | dp = [0 for i in xrange(target + 1)] 17 | dp[0] = 1 18 | 19 | for j in xrange(1, target+1): 20 | for a in nums: 21 | if j >= a: 22 | dp[j] += dp[j - a] 23 | 24 | return dp[target] 25 | ``` 26 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/backpack.md: -------------------------------------------------------------------------------- 1 | # 背包问题 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/backpack/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

在n个物品中挑选若干物品装入背包，最多能装多满？假设背包的大小为m，每个物品的大小为A[i]

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: [3,4,8,5], backpack size=10 9 | 输出: 9 10 | 11 | 样例 2: 12 | 输入: [2,3,5,7], backpack size=12 13 | 输出: 12 14 | 15 | ``` 16 | ### 参考代码 17 | 传统的背包问题解法。二维动态规划。 18 | ```python 19 | class Solution: 20 | """ 21 | @param m: An integer m denotes the size of a backpack 22 | @param A: Given n items with size A[i] 23 | @return: The maximum size 24 | """ 25 | def backPack(self, m, A): 26 | n = len(A) 27 | f = [[False] * (m + 1) for _ in range(n + 1)] 28 | 29 | f[0][0] = True 30 | for i in range(1, n + 1): 31 | f[i][0] = True 32 | for j in range(1, m + 1): 33 | if j >= A[i - 1]: 34 | f[i][j] = f[i - 1][j] or f[i - 1][j - A[i - 1]] 35 | else: 36 | f[i][j] = f[i - 1][j] 37 | 38 | for i in range(m, -1, -1): 39 | if f[n][i]: 40 | return i 41 | return 0 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/best-time-to-buy-and-sell-stock-ii.md: -------------------------------------------------------------------------------- 1 | # 买卖股票的最佳时机 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/best-time-to-buy-and-sell-stock-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个数组 `prices` 表示一支股票每天的价格. 5 | 6 | 你可以完成任意次数的交易, 不过你不能同时参与多个交易 (也就是说, 如果你已经持有这支股票, 在再次购买之前, 你必须先卖掉它). 7 | 8 | 设计一个算法求出最大的利润. 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: [2, 1, 2, 0, 1] 14 | 输出: 2 15 | 解释: 16 | 1. 在第 2 天以 1 的价格买入, 然后在第 3 天以 2 的价格卖出, 利润 1 17 | 2. 在第 4 天以 0 的价格买入, 然后在第 5 天以 1 的价格卖出, 利润 1 18 | 总利润 2. 19 | ``` 20 | 21 | **样例 2:** 22 | 23 | ``` 24 | 输入: [4, 3, 2, 1] 25 | 输出: 0 26 | 解释: 不进行任何交易, 利润为0. 27 | ``` 28 | ### 参考代码 29 | 贪心法: 只要相邻的两天股票的价格是上升的, 我们就进行一次交易, 获得一定利润. 30 | 31 | 这样的策略不一定是最小的交易次数, 但是一定会得到最大的利润. 32 | ```java 33 | public class Solution { 34 | public int maxProfit(int[] prices) { 35 | int profit = 0; 36 | for (int i = 0; i < prices.length - 1; i++) { 37 | int diff = prices[i + 1] - prices[i]; 38 | if (diff > 0) { 39 | profit += diff; 40 | } 41 | } 42 | return profit; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/binary-tree-leaf-sum.md: -------------------------------------------------------------------------------- 1 | # 二叉树叶子节点之和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/binary-tree-leaf-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算二叉树的叶子节点之和 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入：{1,2,3,4} 9 | 输出：7 10 | 解释： 11 | 1 12 | / \ 13 | 2 3 14 | / 15 | 4 16 | 3+4=7 17 | ``` 18 | **样例 2:** 19 | ``` 20 | 输入：{1,#,3} 21 | 输出：3 22 | 解释： 23 | 1 24 | \ 25 | 3 26 | ``` 27 | ### 参考代码 28 | 递归求和即可 29 | ```python 30 | """ 31 | Definition of TreeNode: 32 | class TreeNode: 33 | def __init__(self, val): 34 | this.val = val 35 | this.left, this.right = None, None 36 | """ 37 | class Solution: 38 | # @param {TreeNode} root the root of the binary tree 39 | # @return {int} an integer 40 | def leafSum(self, root): 41 | # Write your code here 42 | p = [] 43 | self.dfs(root, p) 44 | return sum(p) 45 | def dfs(self, root, p): 46 | if root is None: 47 | return 48 | if root.left is None and root.right is None: 49 | p.append(root.val) 50 | self.dfs(root.left, p) 51 | self.dfs(root.right, p) 52 | ``` 53 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/binary-tree-maximum-path-sum-ii.md: -------------------------------------------------------------------------------- 1 | # 二叉树的最大路径和 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/binary-tree-maximum-path-sum-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一棵二叉树，找出从根节点出发的路径中，和最大的一条。 5 | 6 | 这条路径可以在任何二叉树中的节点结束，但是必须包含至少一个点（也就是根了）。 7 | 8 | ### 样例说明 9 | **样例 1:** 10 | 11 | ``` 12 | 输入: {1,2,3} 13 | 输出: 4 14 | 解释: 15 | 1 16 | / \ 17 | 2 3 18 | 1+3=4 19 | ``` 20 | 21 | **样例 2:** 22 | 23 | ``` 24 | 输入: {1,-1,-1} 25 | 输出: 1 26 | 解释: 27 | 1 28 | / \ 29 | -1 -1 30 | ``` 31 | ### 参考代码 32 | 如果从动态规划的角度思考, 可以是这样的: 33 | 34 | - 状态: f[i] 表示从节点i往下走得到的最大路径和 35 | - 状态转移方程: f[i] = max{f[lc[i]], f[rc[i]], 0} + val[i] 36 | 37 | 不过这道题目是二叉树, 比经典的数字三角形要简单, 无需记忆化/定义额外空间储存答案. 38 | 39 | 一次递归遍历整棵二叉树即可. 40 | ```java 41 | public class Solution { 42 | /** 43 | * @param root the root of binary tree. 44 | * @return an integer 45 | */ 46 | public int maxPathSum2(TreeNode root) { 47 | if (root == null) { 48 | return Integer.MIN_VALUE; 49 | } 50 | 51 | int left = maxPathSum2(root.left); 52 | int right = maxPathSum2(root.right); 53 | return root.val + Math.max(0, Math.max(left, right)); 54 | } 55 | } 56 | ``` 57 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/bitwise-and-of-numbers-range.md: -------------------------------------------------------------------------------- 1 | # 数字范围内位的与 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/bitwise-and-of-numbers-range/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定范围[m，n]，其中0 <= m <= n <= 2147483647，输出此范围内所有数字的按位AND，包括端值。 5 | 6 | 例如，给定范围[5,7]，输出返回4。 7 | ### 样例说明 8 | **样例1** 9 | 10 | ``` 11 | 输入: m=5, n=7 12 | 输出: 4 13 | ``` 14 | 15 | **样例2** 16 | 17 | ``` 18 | 输入: m=14, n=15 19 | 输出: 14 20 | ``` 21 | ### 参考代码 22 | 本题中，我们需要得到[m,n]所有元素按位与的结果。举个例子，当m=26，n=30时，它们的二进制表示为为： 23 | **11**010　　**11**011　　**11**100　　**11**101　　**11**110 24 | 这个样例的答案是**11**000，易见我们发现我们只需要找到m和n最左边的公共部分即可。 25 | 26 | 每次都将n与n-1按位与，当n的二进制为1010时，1010 & 1001 = 1000，相当于把二进制位的最后一个1去掉了。因此我们不断的做n^n-1的操作，直到n小于m相等即可。 27 | ```java 28 | public class Solution { 29 | /** 30 | * @param m: an Integer 31 | * @param n: an Integer 32 | * @return: the bitwise AND of all numbers in [m,n] 33 | */ 34 | public int rangeBitwiseAnd(int m, int n) { 35 | // Write your code here 36 | while(m < n) { 37 | n = n & (n-1); 38 | } 39 | return n; 40 | 41 | } 42 | } 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/can-place-flowers.md: -------------------------------------------------------------------------------- 1 | # 能否放置花 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/can-place-flowers/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 假设你有一个长花圃，其中有些地块已经被种植了，但是有些地块没有。但是，花不能够在相邻的地块下种植 - 他们会争夺水从而导致两者的死亡。 5 | 6 | 给定一个花圃（用一个包含0和1的数组来表示，其中0代表空，1代表非空），和一个数字**n**，返回n朵新的花在这个花圃上以能否在不违反“无相邻花”的规则种植。 7 | ### 样例说明 8 | **样例1** 9 | ``` 10 | 输入: flowerbed = [1,0,0,0,1], n = 1 11 | 输出: True 12 | ``` 13 | **样例2** 14 | ``` 15 | 输入: flowerbed = [1,0,0,0,1], n = 2 16 | 输出: False 17 | ``` 18 | ### 参考代码 19 | 比较基础的模拟题，按照题目要求从头开始模拟种植即可。 20 | ```java 21 | public class Solution { 22 | public boolean canPlaceFlowers(int[] flowerbed, int n) { 23 | int count = 0; 24 | for (int i = 0; i < flowerbed.length; i++) { 25 | if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] == 0) 26 | && (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) { 27 | flowerbed[i] = 1; 28 | count++; 29 | } 30 | if (count >= n) { 31 | return true; 32 | } 33 | } 34 | return false; 35 | } 36 | } 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/char-to-integer.md: -------------------------------------------------------------------------------- 1 | # 字符转整数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/char-to-integer/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将字符转换为一个整数。你可以假设字符是ASCII码，也就是说转换后的整数在0~255之间。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 'a' 9 | 输出: 97 10 | 11 | 样例解释: 12 | 返回ASCII码中对应的数字. 13 | 14 | 样例 2: 15 | 输入: '%' 16 | 输出: 37 17 | 18 | 样例解释: 19 | 返回ASCII码中对应的数字. 20 | 21 | ``` 22 | ### 参考代码 23 | 类型转换 24 | ```java 25 | public class Solution { 26 | /** 27 | * @param character a character 28 | * @return an integer 29 | */ 30 | public int charToInteger(char character) { 31 | return (int) character; 32 | } 33 | } 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/climbing-stairs-ii.md: -------------------------------------------------------------------------------- 1 | # 爬楼梯 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/climbing-stairs-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 一个小孩爬一个 n 层台阶的楼梯。他可以每次跳 1 步， 2 步 或者 3 步。实现一个方法来统计总共有多少种不同的方式爬到最顶层的台阶。 5 | 6 | ### 样例说明 7 | 8 | ### 参考代码 9 | 本题采用动态规划的思想，每一步的答案可以由上三步转移而来。 10 | 初始状态和状态转移方程如下： 11 | $f_0=1,\ f_1 =1,\ f_2=2,f_n=f_{n-1}+f_{n-2}+f_{n-3}(n \geq 3)$ 12 | ```java 13 | public class Solution { 14 | /** 15 | * @param n an integer 16 | * @return an integer 17 | */ 18 | public int climbStairs2(int n) { 19 | int[] f = new int[n+1]; 20 | f[0] = 1; 21 | for (int i = 0; i <= n; i++) 22 | for (int j = 1; j <= 3; j++) { 23 | if (i >= j) { 24 | f[i] += f[i-j]; 25 | } 26 | } 27 | return f[n]; 28 | } 29 | } 30 | ``` 31 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/closest-number-in-sorted-array.md: -------------------------------------------------------------------------------- 1 | # 排序数组中最接近元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/closest-number-in-sorted-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在一个排好序的数组 *A* 中找到 *i* 使得 *A[i]* 最接近 *target* 5 | ### 样例说明 6 | 样例1： 7 | ``` 8 | 输入：[1, 2, 3] ，target = 2 9 | 输出： 1. 10 | ``` 11 | 样例2： 12 | ``` 13 | 输入：[1, 4, 6]， target = 3 14 | 输出： 1. 15 | ``` 16 | 样例3： 17 | ``` 18 | 输入：[1, 4, 6]，target = 5, 19 | 输出： 1 或 2. 20 | ``` 21 | ### 参考代码 22 | 利用二分查找的思想即可 23 | ```python 24 | class Solution: 25 | # @param {int[]} A an integer array sorted in ascending order 26 | # @param {int} target an integer 27 | # @return {int} an integer 28 | def closestNumber(self, A, target): 29 | if not A: 30 | return -1 31 | start, end = 0, len(A) - 1 32 | while start + 1 < end: 33 | mid = (start + end) / 2 34 | if A[mid] == target: 35 | return mid 36 | elif A[mid] > target: 37 | end = mid 38 | else: 39 | start = mid 40 | if A[end] - target > target - A[start]: 41 | return start 42 | else: 43 | return end 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/coins-in-a-line.md: -------------------------------------------------------------------------------- 1 | # 硬币排成线 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/coins-in-a-line/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 有 `n` 个硬币排成一条线。两个参赛者轮流从右边依次拿走 1 或 2 个硬币，直到没有硬币为止。拿到最后一枚硬币的人获胜。 5 | 6 | 请判定 **先手玩家** 必胜还是必败? 7 | 8 | 若必胜, 返回 `true`, 否则返回 `false`. 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: 1 14 | 输出: true 15 | ``` 16 | 17 | **样例 2:** 18 | 19 | ``` 20 | 输入: 4 21 | 输出: true 22 | 解释: 23 | 先手玩家第一轮拿走一个硬币, 此时还剩三个. 24 | 这时无论后手玩家拿一个还是两个, 下一次先手玩家都可以把剩下的硬币拿完. 25 | ``` 26 | ### 参考代码 27 | 可以证明, 当硬币数目是3的倍数的时候, 先手玩家必败, 否则他必胜. 28 | 29 | 当硬币数目是3的倍数时, 每一轮先手者拿a个, 后手者拿3-a个即可, 后手必胜. 30 | 31 | 若不是3的倍数, 先手者可以拿1或2个, 此时剩余硬币个数就变成了3的倍数. 32 | ```cpp 33 | class Solution { 34 | public: 35 | /** 36 | * @param n: An integer 37 | * @return: A boolean which equals to true if the first player will win 38 | */ 39 | bool firstWillWin(int n) { 40 | return n % 3; 41 | } 42 | }; 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/combination-sum-iv.md: -------------------------------------------------------------------------------- 1 | # 组合总和 IV 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/combination-sum-iv/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一个都是正整数的数组 `nums`，其中没有重复的数。从中找出所有的和为 `target` 的组合个数。 5 | ### 样例说明 6 | **样例1** 7 | 8 | ``` 9 | 输入: nums = [1, 2, 4] 和 target = 4 10 | 输出: 6 11 | 解释: 12 | 可能的所有组合有： 13 | [1, 1, 1, 1] 14 | [1, 1, 2] 15 | [1, 2, 1] 16 | [2, 1, 1] 17 | [2, 2] 18 | [4] 19 | ``` 20 | 21 | **样例2** 22 | 23 | ``` 24 | 输入: nums = [1, 2] 和 target = 4 25 | 输出: 5 26 | 解释: 27 | 可能的所有组合有： 28 | [1, 1, 1, 1] 29 | [1, 1, 2] 30 | [1, 2, 1] 31 | [2, 1, 1] 32 | [2, 2] 33 | ``` 34 | ### 参考代码 35 | ```cpp 36 | class Solution { 37 | public: 38 | int combinationSum4(vector& nums, int target) { 39 | int n = nums.size(); 40 | int f[target+1]; 41 | memset(f, 0, sizeof(f)); 42 | 43 | f[0] = 1; 44 | for (int i = 1; i <= target; i++) { 45 | for (int j = 0; j < n; j++) 46 | if (i - nums[j] >= 0){ 47 | f[i] += f[i-nums[j]]; 48 | } 49 | } 50 | return f[target]; 51 | } 52 | }; 53 | ``` 54 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/compare-strings.md: -------------------------------------------------------------------------------- 1 | # 比较字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/compare-strings/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 比较两个字符串A和B，确定A中是否包含B中所有的字符。字符串A和B中的字符都是 **大写字母** 5 | ### 样例说明 6 | 给出 A = `"ABCD"` B = `"ACD"`，返回 `true` 7 | 8 | 给出 A = `"ABCD"` B = `"AABC"`， 返回 `false` 9 | ### 参考代码 10 | ```python 11 | class Solution: 12 | """ 13 | @param A : A string includes Upper Case letters 14 | @param B : A string includes Upper Case letters 15 | @return : if string A contains all of the characters in B return True else return False 16 | """ 17 | def compareStrings(self, A, B): 18 | # write your code here 19 | if len(B) == 0: 20 | return True 21 | if len(A) == 0: 22 | return False 23 | #trackTable首先记录A中所有的字符以及它们的个数，然后遍历B,如果出现trackTable[i]小于0的情况，说明B中该字符出现的次数 24 | #大于在A中出现的次数 25 | trackTable = [0 for _ in range(26)] 26 | for i in A: 27 | trackTable[ord(i) - 65] += 1 28 | for i in B: 29 | if trackTable[ord(i) - 65] == 0: 30 | return False 31 | else: 32 | trackTable[ord(i) -65] -= 1 33 | return True 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/convert-linked-list-to-array-list.md: -------------------------------------------------------------------------------- 1 | # 链表转数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/convert-linked-list-to-array-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个链表转换为一个数组。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 1->2->3->null 9 | 输出: [1,2,3] 10 | ``` 11 | **样例 2:** 12 | ``` 13 | 输入: 3->5->8->null 14 | 输出: [3,5,8] 15 | ``` 16 | 17 | 18 | ### 参考代码 19 | ```python 20 | """ 21 | Definition of ListNode 22 | class ListNode(object): 23 | 24 | def __init__(self, val, next=None): 25 | self.val = val 26 | self.next = next 27 | """ 28 | class Solution: 29 | """ 30 | @param {ListNode} head the first node of the linked list. 31 | @return {int[]} an integer list 32 | """ 33 | def toArrayList(self, head): 34 | # Write your code here 35 | r = [] 36 | #遍历链表，并将每个节点的val存储在数组中 37 | while head is not None: 38 | r.append(head.val) 39 | head = head.next 40 | return r 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/count-1-in-binary.md: -------------------------------------------------------------------------------- 1 | # 二进制中有多少个1 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-1-in-binary/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算在一个 32 位的整数的二进制表示中有多少个 `1`。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：32 9 | 输出：1 10 | 解释： 11 | 32(100000)，返回 1。 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：5 16 | 输出：2 17 | 解释： 18 | 5(101)，返回 2。 19 | ``` 20 | ### 参考代码 21 | ```java 22 | public class Solution { 23 | /** 24 | * @param num: an integer 25 | * @return: an integer, the number of ones in num 26 | */ 27 | public int countOnes(int num) { 28 | int count = 0; 29 | while (num != 0) { 30 | num = num & (num - 1); 31 | count++; 32 | } 33 | return count; 34 | } 35 | }; 36 | 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/count-characters.md: -------------------------------------------------------------------------------- 1 | # 字符计数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-characters/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 对一个字符串中的字符进行计数, 返回一个hashmap, key为字符, value是这个字符出现的次数. 5 | ### 样例说明 6 | 7 | 例1: 8 | ``` 9 | 输入: 10 | str = "abca" 11 | 12 | 输出: 13 | { 14 | "a": 2, 15 | "b": 1, 16 | "c": 1 17 | } 18 | 19 | ``` 20 | 例2: 21 | ``` 22 | 输入: 23 | str = "ab" 24 | 25 | 输出: 26 | { 27 | "a": 1, 28 | "b": 1 29 | } 30 | 31 | ``` 32 | 33 | ### 参考代码 34 | 利用map统计即可 35 | ```java 36 | public class Solution { 37 | /* 38 | * @param : a string 39 | * @return: a hash map 40 | */ 41 | public Map countCharacters(String str) { 42 | // Write your code here 43 | Map map = new HashMap(); 44 | for (char c : str.toCharArray()) { 45 | if (!map.containsKey(c)) { 46 | map.put(c, 1); 47 | } else { 48 | map.put(c, map.get(c) + 1); 49 | } 50 | } 51 | return map; 52 | } 53 | }; 54 | ``` 55 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/count-linked-list-nodes.md: -------------------------------------------------------------------------------- 1 | # 链表节点计数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/count-linked-list-nodes/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 计算链表中有多少个节点. 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 1->3->5->null 9 | 输出: 3 10 | 11 | 样例解释: 12 | 返回链表中结点个数，也就是链表的长度. 13 | 14 | 样例 2: 15 | 输入: null 16 | 输出: 0 17 | 18 | 样例解释: 19 | 空链表长度为0 20 | 21 | ``` 22 | ### 参考代码 23 | 简单模拟 24 | ```cpp 25 | /** 26 | * Definition of ListNode 27 | * class ListNode { 28 | * public: 29 | * int val; 30 | * ListNode *next; 31 | * ListNode(int val) { 32 | * this->val = val; 33 | * this->next = NULL; 34 | * } 35 | * } 36 | */ 37 | class Solution { 38 | public: 39 | /** 40 | * @param head the first node of linked list. 41 | * @return an integer 42 | */ 43 | int countNodes(ListNode *head) { 44 | // Write your code here 45 | int cnt = 0; 46 | while (head != NULL) { 47 | cnt ++; 48 | head = head->next; 49 | } 50 | return cnt; 51 | } 52 | }; 53 | ``` 54 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/course-schedule-iii.md: -------------------------------------------------------------------------------- 1 | # 课程表 III 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/course-schedule-iii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 这里有 `n` 门不同的线上课程, 编号为 `1` 到 `n`. 每一节课都有持续时间(课程长度) `t` 和 在第 `d` 天关闭. 课程应连续持续 `t` 天，必须在第`d `天或之前完成, 你将从第一天开始. 5 | 给出 `n` 门线上课程用 `pairs (t, d)` 来表示, 你的任务是找到可以上的最大数量的课程数. 6 | ### 样例说明 7 | 8 | ### 参考代码 9 | 贪心法 Greedy 10 | 课程按照 deadline 排序，从左到右扫描每个课程，依次学习。如果发现学了之后超过 deadline 的，就从之前学过的课程里扔掉一个耗时最长的。 11 | 因为这样可以使得其他的课程往前挪，而往前挪是没影响的。 12 | 所以挑最大值这个事情就是 Heap 的事情了 13 | ```python 14 | import heapq 15 | 16 | 17 | class Solution: 18 | """ 19 | @param courses: duration and close day of each course 20 | @return: the maximal number of courses that can be taken 21 | """ 22 | def scheduleCourse(self, courses): 23 | courses = sorted(courses, key=lambda x: x[1]) 24 | curt_time = 0 25 | heap = [] 26 | for duration, deadline in courses: 27 | curt_time += duration 28 | heapq.heappush(heap, -duration) 29 | if curt_time > deadline: 30 | curt_time -= (-heapq.heappop(heap)) 31 | 32 | return len(heap) 33 | ``` 34 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/delete-node-in-the-middle-of-singly-linked-list.md: -------------------------------------------------------------------------------- 1 | # 在O(1)时间复杂度删除链表节点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/delete-node-in-the-middle-of-singly-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个单链表中的一个等待被删除的节点(非表头或表尾)。请在在 O(1) 时间复杂度删除该链表节点。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | ```python 9 | """ 10 | Definition of ListNode 11 | class ListNode(object): 12 | 13 | def __init__(self, val, next=None): 14 | self.val = val 15 | self.next = next 16 | """ 17 | class Solution: 18 | # @param node: the node in the list should be deleted 19 | # @return: nothing 20 | def deleteNode(self, node): 21 | # write your code here 22 | if node is None or node.next is None: 23 | return 24 | next = node.next; 25 | node.val = next.val 26 | node.next = next.next 27 | return 28 | ``` 29 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/drop-eggs.md: -------------------------------------------------------------------------------- 1 | # 丢鸡蛋 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/drop-eggs/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 楼有 `n` 层高，鸡蛋若从 *k* 层或以上扔，就会碎。从 *k* 层以下扔，就不会碎。 5 | 6 | 现在给两个鸡蛋，用最少的扔的次数找到 *k*。返回最坏情况下次数。 7 | ### 样例说明 8 | **样例 1：** 9 | ``` 10 | 输入：100 11 | 输出：14 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：10 16 | 输出：4 17 | ``` 18 | ### 参考代码 19 | ```cpp 20 | class Solution { 21 | public: 22 | /** 23 | * @param n an integer 24 | * @return an integer 25 | */ 26 | int dropEggs(int n) { 27 | // Write your code here 28 | long long ans = 0; 29 | int x = 0; 30 | while (ans < n) { 31 | x += 1; 32 | ans += x; 33 | } 34 | return x; 35 | } 36 | }; 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/excel-sheet-column-title.md: -------------------------------------------------------------------------------- 1 | # Excel表列标题 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/excel-sheet-column-title/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个正整数，返回相应的列标题，如Excel表中所示。 5 | ### 样例说明 6 | **样例1** 7 | 8 | ``` 9 | 输入: 28 10 | 输出: "AB" 11 | ``` 12 | 13 | **样例2** 14 | 15 | ``` 16 | 输入: 29 17 | 输出: "AC" 18 | ``` 19 | ### 参考代码 20 | ```cpp 21 | class Solution { 22 | public: 23 | string convertToTitle(int n) { 24 | if (n == 0) { 25 | return ""; 26 | } 27 | return convertToTitle((n - 1) / 26) + (char)((n - 1) % 26 + 'A'); 28 | } 29 | }; 30 | ``` 31 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/find-node-in-linked-list.md: -------------------------------------------------------------------------------- 1 | # 在链表中找节点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/find-node-in-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在链表中找值为 value 的节点，如果没有的话，返回空(null)。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 1->2->3 and value = 3 9 | 输出: 最后一个结点 10 | ``` 11 | **样例 2:** 12 | ``` 13 | 输入: 1->2->3 and value = 4 14 | 输出: null 15 | ``` 16 | ### 参考代码 17 | 简单模拟，复杂度O(n) 18 | ```java 19 | /** 20 | * Definition for ListNode 21 | * public class ListNode { 22 | * int val; 23 | * ListNode next; 24 | * ListNode(int x) { 25 | * val = x; 26 | * next = null; 27 | * } 28 | * } 29 | */ 30 | public class Solution { 31 | /** 32 | * @param head: the head of linked list. 33 | * @param val: an integer 34 | * @return: a linked node or null 35 | */ 36 | public ListNode findNode(ListNode head, int val) { 37 | for (ListNode node = head; node != null; node = node.next) { 38 | if (node.val == val) { 39 | return node; 40 | } 41 | } 42 | return null; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/first-unique-character-in-a-string.md: -------------------------------------------------------------------------------- 1 | # 第一个只出现一次的字符 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/first-unique-character-in-a-string/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一个字符串，找出第一个只出现一次的字符。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: "abaccdeff" 9 | 输出: 'b' 10 | 11 | 解释: 12 | 'b' 是第一个出现一次的字符 13 | 14 | 15 | 样例 2: 16 | 输入: "aabccd" 17 | 输出: 'b' 18 | 19 | 解释: 20 | 'b' 是第一个出现一次的字符 21 | 22 | 23 | ``` 24 | ### 参考代码 25 | 辅助数组记录字符出现次数。 26 | ```python 27 | class Solution: 28 | """ 29 | @param str: str: the given string 30 | @return: char: the first unique character in a given string 31 | """ 32 | def firstUniqChar(self, str): 33 | counter = {} 34 | 35 | for c in str: 36 | counter[c] = counter.get(c, 0) + 1 37 | 38 | for c in str: 39 | if counter[c] == 1: 40 | return c 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/flip-bits.md: -------------------------------------------------------------------------------- 1 | # 将整数A转换为B 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/flip-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 如果要将整数n转换为m，需要改变多少个bit位？ 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | a ^ b 得到的结果的二进制中，1的个数就是a和b相异的位数。 9 | ```java 10 | class Solution { 11 | /** 12 | *@param a, b: Two integer 13 | *return: An integer 14 | */ 15 | public static int bitSwapRequired(int a, int b) { 16 | // write your code here 17 | int count = 0; 18 | for (int c = a ^ b; c != 0; c = c >>> 1) { 19 | count += c & 1; 20 | } 21 | return count; 22 | } 23 | }; 24 | ``` 25 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/generate-arraylist-with-given-size.md: -------------------------------------------------------------------------------- 1 | # 生成给定大小的数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/generate-arraylist-with-given-size/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你一个大小size,生成一个元素`从1 到 size`的数组 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: size = 4 9 | 输出: [1, 2, 3, 4] 10 | 11 | 样例解释: 12 | 返回一个顺序填充1到4的数组。 13 | 14 | 样例 2: 15 | 输入: size = 1 16 | 输出: [1] 17 | 18 | 样例解释: 19 | 返回一个顺序填充1到1的数组 20 | 21 | ``` 22 | 23 | ### 参考代码 24 | 基本模拟 25 | ```java 26 | public class Solution { 27 | /** 28 | * @param size an integer 29 | * @return an array list 30 | */ 31 | public ArrayList generate(int size) { 32 | // Write your code here 33 | ArrayList result = new ArrayList(); 34 | for (int i = 1; i <= size; ++i) 35 | result.add(i); 36 | return result; 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/h-index-ii.md: -------------------------------------------------------------------------------- 1 | # H指数 II 2 | 3 | 4 | 5 | 6 | 7 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/h-index-ii/?utm_source=sc-github-wzz) 8 | ## 题目描述 9 | 给定一位研究者论文被引用次数的数组（被引用次数是非负整数），数组已经按照升序排列。编写一个方法，计算出研究者的 h 指数。 10 | 11 | h 指数的定义: “h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）" 12 | 13 | ### 样例说明 14 | ``` 15 | 输入: citations = [0,1,3,5,6] 16 | 输出: 3 17 | 解释: 给定数组表示研究者总共有 5 篇论文，每篇论文相应的被引用了 0, 1, 3, 5, 6 次。 18 |   由于研究者有 3 篇论文每篇至少被引用了 3 次，其余两篇论文每篇被引用不多于 3 次，所以她的 h 指数是 3。 19 | ``` 20 | ### 参考代码 21 | ```cpp 22 | class Solution { 23 | public: 24 | int hIndex(vector& citations) { 25 | int ans = 0; 26 | int h; 27 | int l = 0, r = citations.size() - 1; 28 | 29 | while(l <= r) { 30 | int mid = (l + r)/2; 31 | h = citations.size() - mid; 32 | if (citations[mid] >= h) { 33 | ans = max(ans, h); 34 | r = mid - 1; 35 | } else { 36 | ans = max(ans, citations[mid]); 37 | l = mid + 1; 38 | } 39 | } 40 | return ans; 41 | } 42 | }; 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/hamming-distance.md: -------------------------------------------------------------------------------- 1 | # Hamming距离 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/hamming-distance/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 两个整数的Hamming距离是对应比特位不同的个数。 5 | 给定两个整数x和y，计算两者的Hamming距离。 6 | ### 样例说明 7 | **样例1** 8 | 9 | ``` 10 | 输入: x = 1 和 y = 4 11 | 输出: 2 12 | 解释: 13 | 1的二进制表示是001 14 | 4的二进制表示是100 15 | 共有2位不同 16 | ``` 17 | 18 | 19 | 20 | **样例2** 21 | 22 | ``` 23 | 输入: x = 5 和 y = 2 24 | 输出: 3 25 | 解释: 26 | 5的二进制表示是101 27 | 2的二进制表示是010 28 | 共有3位不同 29 | ``` 30 | ### 参考代码 31 | ```java 32 | public class Solution { 33 | public int hammingDistance(int x, int y) { 34 | int Distance=0; 35 | 36 | while ( x != 0 || y != 0 ) { 37 | if ( x % 2 != y % 2 ) { 38 | Distance ++; 39 | } 40 | x /= 2; 41 | y /= 2; 42 | } 43 | return Distance; 44 | } 45 | } 46 | ``` 47 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/house-robber.md: -------------------------------------------------------------------------------- 1 | # 打劫房屋 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/house-robber/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 假设你是一个专业的窃贼，准备沿着一条街打劫房屋。每个房子都存放着特定金额的钱。你面临的唯一约束条件是：相邻的房子装着相互联系的防盗系统，且 **当相邻的两个房子同一天被打劫时，该系统会自动报警**。 5 | 6 | 给定一个非负整数列表，表示每个房子中存放的钱， 算一算，如果今晚去打劫，**在不触动报警装置的情况下**, 你最多可以得到多少钱 。 7 | ### 样例说明 8 | **样例 1:** 9 | 10 | ``` 11 | 输入: [3, 8, 4] 12 | 输出: 8 13 | 解释: 仅仅打劫第二个房子. 14 | ``` 15 | 16 | **样例 2:** 17 | 18 | ``` 19 | 输入: [5, 2, 1, 3] 20 | 输出: 8 21 | 解释: 抢第一个和最后一个房子 22 | ``` 23 | ### 参考代码 24 | 使用滚动数组的做法，将每个下标直接 % 3。 25 | 时间复杂度 $O(n)$，空间复杂度 $O(1)$ 26 | ```python 27 | class Solution: 28 | """ 29 | @param A: An array of non-negative integers 30 | @return: The maximum amount of money you can rob tonight 31 | """ 32 | def houseRobber(self, A): 33 | if not A: 34 | return 0 35 | if len(A) <= 2: 36 | return max(A) 37 | 38 | f = [0] * 3 39 | f[0], f[1] = A[0], max(A[0], A[1]) 40 | 41 | for i in range(2, len(A)): 42 | f[i % 3] = max(f[(i - 1) % 3], f[(i - 2) % 3] + A[i]) 43 | 44 | return f[(len(A) - 1) % 3] 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/insert-node-in-sorted-linked-list.md: -------------------------------------------------------------------------------- 1 | # 在排序链表中插入一个节点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/insert-node-in-sorted-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在链表中插入一个节点。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：head = 1->4->6->8->null, val = 5 9 | 输出：1->4->5->6->8->null 10 | ``` 11 | **样例 2：** 12 | ``` 13 | 输入：head = 1->null, val = 2 14 | 输出：1->2->null 15 | ``` 16 | ### 参考代码 17 | ---------------------------------------------------------- 18 | ```python 19 | """ 20 | Definition of ListNode 21 | class ListNode(object): 22 | 23 | def __init__(self, val, next=None): 24 | self.val = val 25 | self.next = next 26 | """ 27 | 28 | class Solution: 29 | # @param head, a ListNode 30 | # @param val, an integer 31 | # @return the head of new linked list 32 | def insertNode(self, head, val): 33 | # Write your code here 34 | dummy = ListNode(0, head) 35 | p = dummy 36 | while p.next and p.next.val < val: 37 | p = p.next 38 | node = ListNode(val, p.next) 39 | p.next = node 40 | return dummy.next 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/insertion-sort-list.md: -------------------------------------------------------------------------------- 1 | # 链表插入排序 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/insertion-sort-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用插入排序对链表排序 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 0->null 9 | 输出: 0->null 10 | 11 | 12 | 样例 2: 13 | 输入: 1->3->2->0->null 14 | 输出 :0->1->2->3->null 15 | 16 | 17 | ``` 18 | 19 | ### 参考代码 20 | Sort a linked list using insertion sort. 21 | 22 | 与数组普通插入排序一样的做法。 23 | ```java 24 | public class Solution { 25 | public ListNode insertionSortList(ListNode head) { 26 | ListNode dummy = new ListNode(0); 27 | // 这个dummy的作用是，把head开头的链表一个个的插入到dummy开头的链表里 28 | // 所以这里不需要dummy.next = head; 29 | 30 | while (head != null) { 31 | ListNode node = dummy; 32 | while (node.next != null && node.next.val < head.val) { 33 | node = node.next; 34 | } 35 | ListNode temp = head.next; 36 | head.next = node.next; 37 | node.next = head; 38 | head = temp; 39 | } 40 | 41 | return dummy.next; 42 | } 43 | } 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/is-subsequence.md: -------------------------------------------------------------------------------- 1 | # 是子序列吗？ 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/is-subsequence/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定字符串`s`和`t`，判断`s`是否为`t`的子序列。 5 | 6 | 你可以认为在`s`和`t`中都只包含小写字母。`t`可能是一个非常长（`length ~= 500,000`）的字符串，而`s`是一个较短的字符串（`length <= 100`）。 7 | 8 | 一个字符串的子序列是在原字符串中删去一些字符（也可以不删除）后，不改变剩余字符的相对位置形成的新字符串（例如，`"ace"`是`"abcde"`的子序列而`"aec"`不是）。 9 | ### 样例说明 10 | **样例1：** 11 | ``` 12 | 输入：s = "abc"，t = "ahbgdc" 13 | 输出：true 14 | ``` 15 | **样例2：** 16 | ``` 17 | 输入：s = "axc"，t = "ahbgdc" 18 | 输出：false 19 | ``` 20 | ### 参考代码 21 | 直接按s串的顺序去遍历t串即可。 22 | ```java 23 | public class Solution { 24 | public boolean isSubsequence(String s, String t) { 25 | int i = 0, j = 0; 26 | while(i 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/kth-largest-element-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 找到数组中第K大的元素，N远大于K。请注意你需要找的是数组排序后的第 k 个最大的元素，而不是第 k 个不同的元素 5 | ### 样例说明 6 | **例1:** 7 | ``` 8 | 输入:[9,3,2,4,8],3 9 | 输出:4 10 | 11 | ``` 12 | **例2:** 13 | ``` 14 | 输入:[1,2,3,4,5,6,8,9,10,7],10 15 | 输出:1 16 | 17 | ``` 18 | 19 | ### 参考代码 20 | 利用优先队列求第k大元素 21 | ```java 22 | class Solution { 23 | /** 24 | * @param nums an integer unsorted array 25 | * @param k an integer from 1 to n 26 | * @return the kth largest element 27 | */ 28 | public int kthLargestElement2(int[] nums, int k) { 29 | // Write your code here 30 | PriorityQueue q = new PriorityQueue(k); 31 | for (int num : nums) { 32 | q.offer(num); 33 | if (q.size() > k) { 34 | q.poll(); 35 | } 36 | } 37 | return q.peek(); 38 | } 39 | }; 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/linked-list-cycle.md: -------------------------------------------------------------------------------- 1 | # 带环链表 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/linked-list-cycle/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个链表，判断它是否有环。

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 21->10->4->5, then tail connects to node index 1(value 10). 9 | 输出: true 10 | 11 | 样例 2: 12 | 输入: 21->10->4->5->null 13 | 输出: false 14 | 15 | ``` 16 | 17 | ### 参考代码 18 | Given a linked list, determine if it has a cycle in it. 19 | 20 | Follow up: 21 | Can you solve it without using extra space? 22 | 23 | 快慢指针的经典题。 24 | 快指针每次走两步，慢指针一次走一步。 25 | 在慢指针进入环之后，快慢指针之间的距离每次缩小1，所以最终能相遇。 26 | ```java 27 | public class Solution { 28 | public Boolean hasCycle(ListNode head) { 29 | if (head == null || head.next == null) { 30 | return false; 31 | } 32 | 33 | ListNode fast, slow; 34 | fast = head.next; 35 | slow = head; 36 | while (fast != slow) { 37 | if(fast==null || fast.next==null) 38 | return false; 39 | fast = fast.next.next; 40 | slow = slow.next; 41 | } 42 | return true; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/longest-common-subsequence.md: -------------------------------------------------------------------------------- 1 | # 最长公共子序列 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/longest-common-subsequence/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给出两个字符串，找到最长公共子序列(LCS)，返回LCS的长度。

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: "ABCD" and "EDCA" 9 | 输出: 1 10 | 11 | 解释: 12 | LCS 是 'A' 或 'D' 或 'C' 13 | 14 | 15 | 样例 2: 16 | 输入: "ABCD" and "EACB" 17 | 输出: 2 18 | 19 | 解释: 20 | LCS 是 "AC" 21 | ``` 22 | 23 | ### 参考代码 24 | 使用九章算法强化班和动态规划专题班中讲的匹配型动态规划 25 | ```python 26 | class Solution: 27 | """ 28 | @param A: A string 29 | @param B: A string 30 | @return: The length of longest common subsequence of A and B 31 | """ 32 | def longestCommonSubsequence(self, A, B): 33 | n, m = len(A), len(B) 34 | f = [[0] * (m + 1) for i in range(n + 1)] 35 | for i in range(1, n + 1): 36 | for j in range(1, m + 1): 37 | f[i][j] = max(f[i][j - 1], f[i - 1][j]) 38 | if A[i - 1] == B[j - 1]: 39 | f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1) 40 | return f[n][m] 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/longest-words.md: -------------------------------------------------------------------------------- 1 | # 最长单词 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/longest-words/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一个词典，找出其中所有最长的单词。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | ```java 9 | class Solution { 10 | /** 11 | * @param dictionary: an array of strings 12 | * @return: an arraylist of strings 13 | */ 14 | ArrayList longestWords(String[] dictionary) { 15 | // write your code here 16 | int maxLen = 0; 17 | ArrayList ans = new ArrayList(); 18 | for (int i=0; imaxLen) maxLen = dictionary[i].length(); 20 | for (int i=0; i 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/lowercase-to-uppercase-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个字符串中的小写字母转换为大写字母。不是字母的字符不需要做改变。 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: str = "abc" 10 | 输出: "ABC" 11 | ``` 12 | **样例 2:** 13 | 14 | ``` 15 | 输入: str = "aBc" 16 | 输出: "ABC" 17 | ``` 18 | **样例 3:** 19 | 20 | ``` 21 | 输入: str = "abC12" 22 | 输出: "ABC12" 23 | ``` 24 | ### 参考代码 25 | ```java 26 | public class Solution { 27 | /** 28 | * @param str a string 29 | * @return a string 30 | */ 31 | public String lowercaseToUppercase2(String str) { 32 | // Write your code here 33 | StringBuilder sb = new StringBuilder(str); 34 | //遍历整个字符串，将所有的小写字母转成大写字母 35 | for (int index = 0; index < sb.length(); index++) { 36 | char c = sb.charAt(index); 37 | if (Character.isLowerCase(c)) { 38 | sb.setCharAt(index, Character.toUpperCase(c)); 39 | } 40 | } 41 | return sb.toString(); 42 | } 43 | } 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/lowercase-to-uppercase.md: -------------------------------------------------------------------------------- 1 | # 大小写转换 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/lowercase-to-uppercase/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个字符由小写字母转换为大写字母 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: 'a' 10 | 输出: 'A' 11 | ``` 12 | **样例 2:** 13 | 14 | ``` 15 | 输入: 'b' 16 | 输出: 'B' 17 | ``` 18 | ### 参考代码 19 | ```java 20 | public class Solution { 21 | /** 22 | * @param character a character 23 | * @return a character 24 | */ 25 | public char lowercaseToUppercase(char character) { 26 | // Write your code here 27 | //获得character与'a'的差值，在'A'的基础上加上这个差值 28 | return (char)(character - 'a' + 'A'); 29 | } 30 | } 31 | ``` 32 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/majority-number.md: -------------------------------------------------------------------------------- 1 | # 主元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/majority-number/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个整型数组，找出主元素，它在数组中的出现次数严格大于数组元素个数的二分之一。


5 | ### 样例说明 6 | 7 | ### 参考代码 8 |

    

9 | ```java 10 | public class Solution { 11 | /** 12 | * @param nums: a list of integers 13 | * @return: find a majority number 14 | */ 15 | public int majorityNumber(ArrayList nums) { 16 | //由于主元素在数组中个数严格大于1/2，所以假设当前元素就是主元素，用count记录当前元素个数与其他元素个数的差值， 17 | //candidate记录当前元素是什么，最后count一定为正数并且candidate就是主元素 18 | int count = 0, candidate = -1; 19 | for (int i = 0; i < nums.size(); i++) { 20 | if (count == 0) { 21 | candidate = nums.get(i); 22 | count = 1; 23 | } else if (candidate == nums.get(i)) { 24 | count++; 25 | } else { 26 | count--; 27 | } 28 | } 29 | return candidate; 30 | } 31 | } 32 | ``` 33 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/max-of-3-numbers.md: -------------------------------------------------------------------------------- 1 | # 三数之中的最大值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/max-of-3-numbers/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给三个整数，求他们中的最大值。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: num1 = 1, num2 = 9, num3 = 0 9 | 输出: 9 10 | 11 | 样例解释: 12 | 返回三个数中最大的数。 13 | 14 | 样例 2: 15 | 输入: num1 = 1, num2 = 2, num3 = 3 16 | 输出: 3 17 | 18 | 样例解释: 19 | 返回三个中最大的数字。 20 | 21 | ``` 22 | ### 参考代码 23 | if选择语句的基本操作 24 | ```java 25 | public class Solution { 26 | /** 27 | * @param a an integer 28 | * @param b an integer 29 | * @param c an integer 30 | * @return an integer 31 | */ 32 | public int maxOfThreeNumbers(int a, int b, int c) { 33 | if (a >= b && a >= c) { 34 | return a; 35 | } else if (b >= a && b >= c) { 36 | return b; 37 | } else { 38 | return c; 39 | } 40 | } 41 | } 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/max-of-array.md: -------------------------------------------------------------------------------- 1 | # 数组的最大值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/max-of-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一个浮点数数组，求数组中的最大值。 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: [1.0, 2.1, -3.3] 10 | 输出: 2.1 11 | 样例解释: 返回最大的数字 12 | ``` 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: [1.0, 1.0, -3.3] 17 | 输出: 1.0 18 | 样例解释: 返回最大的数字。 19 | ``` 20 | ### 参考代码 21 | 模拟过程 22 | ```java 23 | public class Solution { 24 | /** 25 | * @param A a float array 26 | * @return a float number 27 | */ 28 | public float maxOfArray(float[] A) { 29 | float max = A[0]; 30 | for (int i = 1; i < A.length; i++) { 31 | if (A[i] > max) { 32 | max = A[i]; 33 | } 34 | } 35 | return max; 36 | } 37 | } 38 | ``` 39 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/maximum-average-subarray.md: -------------------------------------------------------------------------------- 1 | # 子数组的最大平均值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/maximum-average-subarray/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个由`n`个整数组成的数组，找到给定长度`k`的连续子数组，该子数组具有最大平均值。你需要输出最大平均值。 5 | ### 样例说明 6 | **样例1** 7 | 8 | ``` 9 | 输入: nums = [1,12,-5,-6,50,3] and k = 4 10 | 输出: 12.75 11 | 解释: 12 | 最大平均为(12-5-6+50)/4 = 51/4 = 12.75 13 | ``` 14 | 15 | 16 | 17 | **样例2** 18 | 19 | ``` 20 | 输入: nums = [4,2,1,3,3] and k = 2 21 | 输出: 3.00 22 | 解释: 23 | 最大平均为(3+3)/2 = 6/2 = 3.00 24 | ``` 25 | ### 参考代码 26 | ```java 27 | public class Solution { 28 | /** 29 | * @param nums: an array 30 | * @param k: an integer 31 | * @return: the maximum average value 32 | */ 33 | public double findMaxAverage(int[] nums, int k) { 34 | // Write your code here 35 | int n = nums.length; 36 | int[] sum = new int[n + 1]; 37 | for(int i = 1; i <= n; i++){ 38 | sum[i] = sum[i - 1] + nums[i - 1]; 39 | } 40 | int ans = Integer.MIN_VALUE; 41 | for(int i = k; i <= n; i++){ 42 | ans = Math.max(ans, sum[i] - sum[i - k]); 43 | } 44 | return ans * 1.0 / k; 45 | } 46 | } 47 | ``` 48 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/maximum-subarray.md: -------------------------------------------------------------------------------- 1 | # 最大子数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/maximum-subarray/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组，找到一个具有最大和的子数组，返回其最大和。 5 | ### 样例说明 6 | **样例1:** 7 | 8 | ``` 9 | 输入：[−2,2,−3,4,−1,2,1,−5,3] 10 | 输出：6 11 | 解释：符合要求的子数组为[4,−1,2,1]，其最大和为 6。 12 | ``` 13 | 14 | **样例2:** 15 | 16 | ``` 17 | 输入：[1,2,3,4] 18 | 输出：10 19 | 解释：符合要求的子数组为[1,2,3,4]，其最大和为 10。 20 | ``` 21 | ### 参考代码 22 | 使用前缀和的方法，计算每个位置为结尾的 subarray 的最大值是多少。 23 | ```python 24 | class Solution: 25 | """ 26 | @param nums: A list of integers 27 | @return: A integer indicate the sum of max subarray 28 | """ 29 | def maxSubArray(self, nums): 30 | #prefix_sum记录前i个数的和，max_Sum记录全局最大值，min_Sum记录前i个数中0-k的最小值 31 | min_sum, max_sum = 0, -sys.maxsize 32 | prefix_sum = 0 33 | 34 | for num in nums: 35 | prefix_sum += num 36 | max_sum = max(max_sum, prefix_sum - min_sum) 37 | min_sum = min(min_sum, prefix_sum) 38 | 39 | return max_sum 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/merge-sorted-array-ii.md: -------------------------------------------------------------------------------- 1 | # 合并排序数组 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/merge-sorted-array-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

合并两个有序升序的整数数组A和B变成一个新的数组。新数组也要有序。

5 | ### 样例说明 6 | 7 | ### 参考代码 8 | ```cpp 9 | class Solution { 10 | public: 11 | /** 12 | * @param A and B: sorted integer array A and B. 13 | * @return: A new sorted integer array 14 | */ 15 | vector mergeSortedArray(vector &A, vector &B) { 16 | vector C; 17 | int i = 0, j = 0; 18 | while (i < A.size() && j < B.size()) { 19 | if (A[i] < B[j]) { 20 | C.push_back(A[i++]); 21 | } else { 22 | C.push_back(B[j++]); 23 | } 24 | } 25 | while (i < A.size()) { 26 | C.push_back(A[i++]); 27 | } 28 | while (j < B.size()) { 29 | C.push_back(B[j++]); 30 | } 31 | return C; 32 | } 33 | }; 34 | 35 | ``` 36 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/middle-of-linked-list.md: -------------------------------------------------------------------------------- 1 | # 链表的中点 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/middle-of-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 找链表的中点。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 1->2->3 9 | 输出: 2 10 | 样例解释: 返回中间节点的值 11 | ``` 12 | **样例 2:** 13 | ``` 14 | 输入: 1->2 15 | 输出: 1 16 | 样例解释: 如果长度是偶数，则返回中间偏左的节点的值。 17 | ``` 18 | ### 参考代码 19 | 这里使用快慢指针的方法来寻找终点。 20 | 可以确保时间复杂度是 O(n/2) 21 | 22 | 快指针每次两步，慢指针每次一步。 23 | 快指针到尾的时候，慢指针就是中点。 24 | ```python 25 | """ 26 | Definition of ListNode 27 | class ListNode(object): 28 | 29 | def __init__(self, val, next=None): 30 | self.val = val 31 | self.next = next 32 | """ 33 | 34 | class Solution: 35 | # @param head: the head of linked list. 36 | # @return: a middle node of the linked list 37 | def middleNode(self, head): 38 | # Write your code here 39 | if head is None: 40 | return None 41 | slow = head 42 | fast = slow.next 43 | while fast is not None and fast.next is not None: 44 | slow = slow.next 45 | fast = fast.next.next 46 | 47 | return slow 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/minimum-moves-to-equal-array-elements.md: -------------------------------------------------------------------------------- 1 | # 使数组元素相同的最少步数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/minimum-moves-to-equal-array-elements/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个大小为`n`的**非空**整数数组，找出使得数组中所有元素相同的最少步数，其中一步被定义为将数组中`n - 1`个元素加一。 5 | ### 样例说明 6 | ``` 7 | 输入： 8 | [1,2,3] 9 | 10 | 输出： 11 | 3 12 | 13 | 说明： 14 | 只需要三步即可（每一步将其中两个元素加一）： 15 | 16 | [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4] 17 | ``` 18 | ### 参考代码 19 | ```cpp 20 | class Solution { 21 | public: 22 | int minMoves(vector& nums) { 23 | int sumNum = 0; 24 | int minNum = INT_MAX; 25 | for (int num : nums) { 26 | sumNum += num; 27 | minNum = min(minNum, num); 28 | } 29 | return sumNum - minNum * nums.size(); 30 | } 31 | }; 32 | ``` 33 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/moving-average-from-data-stream.md: -------------------------------------------------------------------------------- 1 | # 数据流滑动窗口平均值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/moving-average-from-data-stream/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一串整数流和窗口大小，计算滑动窗口中所有整数的平均值。 5 | ### 样例说明 6 | **样例1 :** 7 | ``` 8 | MovingAverage m = new MovingAverage(3); 9 | m.next(1) = 1 // 返回 1.00000 10 | m.next(10) = (1 + 10) / 2 // 返回 5.50000 11 | m.next(3) = (1 + 10 + 3) / 3 // 返回 4.66667 12 | m.next(5) = (10 + 3 + 5) / 3 // 返回 6.00000 13 | ``` 14 | ### 参考代码 15 | 使用队列 16 | ```java 17 | public class MovingAverage { 18 | 19 | private Queue que; 20 | private double sum = 0; 21 | private int size; 22 | 23 | /** Initialize your data structure here. */ 24 | public MovingAverage(int size) { 25 | que = new LinkedList(); 26 | this.size = size; 27 | } 28 | 29 | public double next(int val) { 30 | // Write your code here 31 | sum += val; 32 | if (que.size() == size) { 33 | sum = sum - que.poll(); 34 | } 35 | que.offer(val); 36 | return sum / que.size(); 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/name-deduplication.md: -------------------------------------------------------------------------------- 1 | # 姓名去重 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/name-deduplication/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一串名字，将他们去重之后返回。两个名字重复是说在忽略大小写的情况下是一样的。 5 | ### 样例说明 6 | 例1: 7 | ``` 8 | 输入:["James", "james", "Bill Gates", "bill Gates", "Hello World", "HELLO WORLD", "Helloworld"] 9 | 10 | 11 | 输出:["james", "bill gates", "hello world", "helloworld"] 12 | ``` 13 | 例2: 14 | ``` 15 | 输入:["cmy","Cmy"] 16 | 17 | 输出:["cmy"] 18 | ``` 19 | ### 参考代码 20 | 利用map去重 21 | ```cpp 22 | #include 23 | class Solution { 24 | public: 25 | /** 26 | * @param names a string array 27 | * @return a string array 28 | */ 29 | vector nameDeduplication(vector& names) { 30 | // Write your code here 31 | map mp; 32 | vector result; 33 | for (auto& str : names) { 34 | std::transform(str.begin(), str.end(), str.begin(), ::tolower); 35 | if (mp.find(str) == mp.end()) { 36 | mp[str] = true; 37 | result.push_back(str); 38 | } 39 | } 40 | return result; 41 | } 42 | }; 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/next-permutation.md: -------------------------------------------------------------------------------- 1 | # 下一个排列 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/next-permutation/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组来表示排列，找出其之后的一个排列。 5 | ### 样例说明 6 | 例1: 7 | ``` 8 | 输入:[1] 9 | 输出:[1] 10 | ``` 11 | 12 | 13 | 例2: 14 | ``` 15 | 输入:[1,3,2,3] 16 | 输出:[1,3,3,2] 17 | ``` 18 | 19 | 例3: 20 | ``` 21 | 输入:[4,3,2,1] 22 | 输出:[1,2,3,4] 23 | ``` 24 | 25 | ### 参考代码 26 | 从最后一个位置开始，找到一个上升点，上升点之前的无需改动。 27 | 然后，翻转上升点之后的降序。 28 | 在降序里，找到第一个比上升点大的，交换位置。 29 | ```python 30 | class Solution: 31 | # @param num : a list of integer 32 | # @return : a list of integer 33 | def nextPermutation(self, num): 34 | # write your code here 35 | for i in range(len(num)-2, -1, -1): 36 | if num[i] < num[i+1]: 37 | break 38 | else: 39 | num.reverse() 40 | return num 41 | for j in range(len(num)-1, i, -1): 42 | if num[j] > num[i]: 43 | num[i], num[j] = num[j], num[i] 44 | break 45 | for j in range(0, (len(num) - i)//2): 46 | num[i+j+1], num[len(num)-j-1] = num[len(num)-j-1], num[i+j+1] 47 | return num 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/null-terminated-string.md: -------------------------------------------------------------------------------- 1 | # NULL 结尾的字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/null-terminated-string/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用 C 或 C++ 设计一个 void reverse `(char* str)` 功能，实现把一个以 NULL 结尾字符串翻转的功能。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | 首尾双指针，依次交换，直至相遇。 9 | ```cpp 10 | class Solution { 11 | public: 12 | /** 13 | * @param str C-String 14 | * @return void 15 | */ 16 | void reverse(char *str) { 17 | // Write your code here 18 | char * end = str; 19 | char tmp; 20 | if (str) { 21 | while (*end != NULL) { 22 | ++end; 23 | } 24 | } 25 | --end; 26 | while (str < end) { 27 | tmp = *str; 28 | *str++ = *end; 29 | *end-- = tmp; 30 | } 31 | } 32 | }; 33 | ``` 34 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/number-of-1-bits.md: -------------------------------------------------------------------------------- 1 | # 判断一个整数中有多少个1 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/number-of-1-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 写一个函数，其以无符号整数为输入，而输出对应二进制数所具有的“1”的位数（也被称为汉明权重） 5 | ### 样例说明 6 | **样例 1** 7 | ``` 8 | 输入：n = 11 9 | 输出：3 10 | 解析：11(10) = 1011(2), 返回 3 11 | ``` 12 | **样例 2** 13 | ``` 14 | 输入：n = 7 15 | 输出：3 16 | 解析：7(10) = 111(2), 返回 3 17 | ``` 18 | ### 参考代码 19 | 使用位运算符号‘&’，与1做与操作 20 | 并将n右移一位( >>1 ) 21 | 直到n为0 22 | ```java 23 | public class Solution { 24 | /** 25 | * @param n: an unsigned integer 26 | * @return: the number of ’1' bits 27 | */ 28 | public int hammingWeight(int n) { 29 | int ones = 0; 30 | while(n!=0) { 31 | ones += (n & 1); 32 | n = n>>1; 33 | } 34 | return ones; 35 | } 36 | } 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/number-of-ways-to-represent-n-cents.md: -------------------------------------------------------------------------------- 1 | # 凑 N 分钱的方案数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/number-of-ways-to-represent-n-cents/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你无限多个的 25 分，10 分，5 分和 1 分的硬币。问如果要凑出 `n` 分钱有多少种不同的方式？ 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入:n = `11` 9 | 输出: 4 10 | 解释: 11 | 11 = 1 + 1 + 1... + 1 12 | = 1 + 1 + 1 + 1 + 1 + 1 + 5 13 | = 1 + 5 + 5 14 | = 1 + 10 15 | ``` 16 | 17 | **样例 2:** 18 | ``` 19 | 输入:n = `9` 20 | 输出: 2 21 | 解释: 22 | 9 = 1 + 1 + 1... + 1 23 | = 1 + 1 + 1 + 1 + 5 24 | ``` 25 | ### 参考代码 26 | ```java 27 | public class Solution { 28 | /** 29 | * @param n an integer 30 | * @return an integer 31 | */ 32 | public int waysNCents(int n) { 33 | int[] f = new int[n+1]; 34 | f[0] = 1; 35 | int[] cents = new int[]{1, 5, 10, 25}; 36 | for (int i = 0; i < 4; i++) 37 | for (int j = 1; j <= n; j++) { 38 | if (j >= cents[i]) { 39 | f[j] += f[j-cents[i]]; 40 | } 41 | } 42 | return f[n]; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/o1-check-power-of-2.md: -------------------------------------------------------------------------------- 1 | # O(1)时间检测2的幂次 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/o1-check-power-of-2/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用 O(*1*) 时间检测整数 *n* 是否是 *2* 的幂次。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | 2的幂的数字，其二进制中只有一个1。 9 | $n&(n-1) == 0$ 说明只有1个1。 10 | ```cpp 11 | class Solution { 12 | public: 13 | /* 14 | * @param n: An integer 15 | * @return: True or false 16 | */ 17 | bool checkPowerOf2(int n) { 18 | // write your code here 19 | if (n==0) 20 | return false; 21 | while (n % 2 == 0) 22 | n /=2 ; 23 | return n == 1; 24 | } 25 | }; 26 | ``` 27 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/pass-interview.md: -------------------------------------------------------------------------------- 1 | # 通过面试 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/pass-interview/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出面试者的性别和面试成绩，判断求职者是否通过面试。 5 | 6 | * 如果面试者是男性，那么面试成绩必须大于等于4分。 7 | * 如果面试者是女性，那么面试成绩必须大于等于3分。 8 | 9 | 参数sex是boolean类型，sex为true时，代表求职者是女性，因为女人永远都是对的。相反sex为false时，代表求职者为男性，因为男人永远都是错的。 10 | ### 样例说明 11 | 12 | ### 参考代码 13 | ```java 14 | public class Solution { 15 | /** 16 | * @param sex the sex of a candidate 17 | * @param score the interview score of a candidate 18 | * @return a boolean 19 | */ 20 | public boolean passInterview(boolean sex, double score) { 21 | if (sex) { 22 | return score >= 3; 23 | } 24 | return score >= 4; 25 | } 26 | } 27 | 28 | ``` 29 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/path-sum.md: -------------------------------------------------------------------------------- 1 | # 路径和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/path-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定二叉树和求和，确定树是否具有根到叶路径，使得沿路径的所有值相加等于给定的总和。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | 使用深度搜索的方法，当搜索到叶子节点时，比较与目标和的值，若相同，则结果为真，若搜索完毕没有出现目标和，返回假 9 | ```java 10 | /** 11 | * Definition of TreeNode: 12 | * public class TreeNode { 13 | * public int val; 14 | * public TreeNode left, right; 15 | * public TreeNode(int val) { 16 | * this.val = val; 17 | * this.left = this.right = null; 18 | * } 19 | * } 20 | */ 21 | 22 | public class Solution { 23 | /** 24 | * @param root: the tree 25 | * @param sum: the sum 26 | * @return: if the tree has a root-to-leaf path 27 | */ 28 | public boolean pathSum(TreeNode root, int sum) { 29 | // Write your code here. 30 | if (root == null) return false; 31 | else if (root.val == sum && root.left == null && root.right == null) return true; 32 | else return pathSum(root.left, sum - root.val) || pathSum(root.right, sum - root.val); 33 | } 34 | } 35 | ``` 36 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/previous-permutation.md: -------------------------------------------------------------------------------- 1 | # 上一个排列 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/previous-permutation/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组来表示排列，找出其上一个排列。 5 | ### 样例说明 6 | 7 | 例1: 8 | ``` 9 | 输入:[1] 10 | 输出:[1] 11 | ``` 12 | 13 | 14 | 例2: 15 | ``` 16 | 输入:[1,3,2,3] 17 | 输出:[1,2,3,3] 18 | ``` 19 | 20 | 例3: 21 | ``` 22 | 输入:[1,2,3,4] 23 | 输出:[4,3,2,1] 24 | 25 | ``` 26 | 27 | ### 参考代码 28 | 从排列的最末尾开始，找到第一个下降点，下降点的意义为这个点之前的序列无需改动。 29 | 然后,将后面的序列变为降序。 30 | 从下降点开始扫描，找到第一个比她小的数字，交换即可。 31 | ```python 32 | class Solution: 33 | # @param num : a list of integer 34 | # @return : a list of integer 35 | def previousPermuation(self, num): 36 | # write your code here 37 | for i in range(len(num)-2, -1, -1): 38 | if num[i] > num[i+1]: 39 | break 40 | else: 41 | num.reverse() 42 | return num 43 | for j in range(len(num)-1, i, -1): 44 | if num[j] < num[i]: 45 | num[i], num[j] = num[j], num[i] 46 | break 47 | for j in range(0, (len(num) - i)//2): 48 | num[i+j+1], num[len(num)-j-1] = num[len(num)-j-1], num[i+j+1] 49 | return num 50 | ``` 51 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/prime-factorization.md: -------------------------------------------------------------------------------- 1 | # 分解质因数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/prime-factorization/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个整数分解为若干质因数之乘积。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：10 9 | 输出：[2, 5] 10 | ``` 11 | **样例 2：** 12 | ``` 13 | 输入：660 14 | 输出：[2, 2, 3, 5, 11] 15 | ``` 16 | ### 参考代码 17 | ```java 18 | public class Solution { 19 | /** 20 | * @param num an integer 21 | * @return an integer array 22 | */ 23 | public List primeFactorization(int num) { 24 | List factors = new ArrayList(); 25 | 26 | for (int i = 2; i * i <= num; i++) { 27 | while (num % i == 0) { 28 | num = num / i; 29 | factors.add(i); 30 | } 31 | } 32 | 33 | if (num != 1) { 34 | factors.add(num); 35 | } 36 | 37 | return factors; 38 | } 39 | } 40 | 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/print-numbers-by-recursion.md: -------------------------------------------------------------------------------- 1 | # 用递归打印数字 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/print-numbers-by-recursion/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用递归的方法找到从1到最大的N位整数。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入 : N = 1 9 | 输出 :[1,2,3,4,5,6,7,8,9] 10 | ``` 11 | **样例 2:** 12 | ``` 13 | 输入 : N = 2 14 | 输出 :[[1,2,3,4,5,6,7,8,9,10,11,12,...,99] 15 | ``` 16 | 17 | ### 参考代码 18 | ```java 19 | public class Solution { 20 | /** 21 | * @param n: An integer. 22 | * return : An array storing 1 to the largest number with n digits. 23 | */ 24 | public List numbersByRecursion(int n) { 25 | // write your code here 26 | ArrayList res = new ArrayList<>(); 27 | num(n, 0, res); 28 | return res; 29 | } 30 | 31 | public void num(int n, int ans,ArrayList res){ 32 | 33 | if(n == 0){ 34 | if(ans > 0){ 35 | res.add(ans); 36 | } 37 | return; 38 | } 39 | 40 | int i; 41 | for(i = 0; i <= 9; i++){ 42 | num(n - 1, ans * 10 + i, res); 43 | } 44 | 45 | } 46 | } 47 | ``` 48 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/recover-rotated-sorted-array.md: -------------------------------------------------------------------------------- 1 | # 恢复旋转排序数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/recover-rotated-sorted-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个**旋转**排序数组，在原地恢复其排序。（升序） 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | Given a rotated sorted array, recover it to sorted array in-place. 9 | 10 | Example 11 | [4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5] 12 | ```python 13 | class Solution: 14 | """ 15 | @param nums: The rotated sorted array 16 | @return: The recovered sorted array 17 | """ 18 | def recoverRotatedSortedArray(self, nums): 19 | nums.sort() 20 | ``` 21 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/reference.md: -------------------------------------------------------------------------------- 1 | # 引用 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reference/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 实现一个类 *ReferenceManager* 包含如下两种方法 5 | 6 | 1.`copyValue(Node obj)` 只拷贝参数*obj*的权值，*obj*和*node*仍然是两个指针 7 | 8 | 2.`copyReference(Node obj)` *obj*和*node*指向同一个地方 9 | ### 样例说明 10 | ``` 11 | ReferenceManager ref = ReferenceManager(); 12 | Node obj = new Node(0); 13 | ref.copyValue(obj); 14 | ref.node.val; // 值为0 15 | ref.node; // 与obj不同. 16 | 17 | Node obj2 = new Node(1); 18 | ref.copyReference(obj2); 19 | ref.node.val; // 值为1 20 | ref.node; // 与 obj2相同，指向同一个地址 21 | ``` 22 | ### 参考代码 23 | java中直接赋值是同一个变量，指向同一个地址 24 | ```java 25 | public class ReferenceManager { 26 | public Node node; 27 | 28 | public void copyValue(Node obj) { 29 | if (obj == null) { 30 | return; 31 | } 32 | if (node == null) { 33 | node = new Node(obj.val); 34 | } 35 | node.val = obj.val; 36 | } 37 | 38 | public void copyReference(Node obj) { 39 | node = obj; 40 | } 41 | } 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/remove-duplicates-from-sorted-list.md: -------------------------------------------------------------------------------- 1 | # 删除排序链表中的重复元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/remove-duplicates-from-sorted-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个排序链表，删除所有重复的元素每个元素只留下一个。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: null 9 | 输出: null 10 | 11 | 12 | 样例 2: 13 | 输入: 1->1->2->null 14 | 输出: 1->2->null 15 | 16 | 样例 3: 17 | 输入: 1->1->2->3->3->null 18 | 输出: 1->2->3->null 19 | 20 | 21 | ``` 22 | ### 参考代码 23 | Given a sorted linked list, delete all duplicates such that each element appear only once. 24 | 25 | For example, 26 | Given 1->1->2, return 1->2. 27 | Given 1->1->2->3->3, return 1->2->3. 28 | 29 | 跳过相同的元素，指向再后一个元素。 30 | ```java 31 | public class Solution { 32 | public ListNode deleteDuplicates(ListNode head) { 33 | if (head == null) { 34 | return null; 35 | } 36 | 37 | ListNode node = head; 38 | while (node.next != null) { 39 | if (node.val == node.next.val) { 40 | node.next = node.next.next; 41 | } else { 42 | node = node.next; 43 | } 44 | } 45 | return head; 46 | } 47 | } 48 | ``` 49 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/remove-element.md: -------------------------------------------------------------------------------- 1 | # 删除元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/remove-element/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个数组和一个值，在原地删除与值相同的数字，返回新数组的长度。

元素的顺序可以改变，并且对新的数组不会有影响。

5 | ### 样例说明 6 | 7 | ### 参考代码 8 | Given an array and a value, remove all instances of that value in place and return the new length. 9 | 10 | The order of elements can be changed. It doesn't matter what you leave beyond the new length. 11 | 12 | 从前往后，遇到等于value的就跳过，把后面的元素填到前面的位置里。 13 | ```java 14 | public class Solution { 15 | public int removeElement(int[] A, int elem) { 16 | int i = 0; 17 | int pointer = A.length - 1; 18 | while(i <= pointer){ 19 | if(A[i] == elem){ 20 | A[i] = A[pointer]; 21 | pointer--; 22 | } else { 23 | i++; 24 | } 25 | } 26 | return pointer + 1; 27 | } 28 | } 29 | ``` 30 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/reverse-3-digit-integer.md: -------------------------------------------------------------------------------- 1 | # 反转一个3位整数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reverse-3-digit-integer/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 反转一个只有3位数的整数。 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: number = 123 10 | 输出: 321 11 | ``` 12 | **样例 2:** 13 | 14 | ``` 15 | 输入: number = 900 16 | 输出: 9 17 | ``` 18 | 19 | ### 参考代码 20 | 硅谷求职算法集训营上课版本 21 | ```java 22 | public class Solution { 23 | /* 24 | * @param number: A 3-digit number. 25 | * @return: Reversed number. 26 | */ 27 | public int reverseInteger(int number) { 28 | // write your code here 29 | //获得个位数 30 | int num1 = number % 10; 31 | //获得十位数 32 | int num2 = (number / 10) % 10; 33 | //获得百位数 34 | int num3 = ((number / 10) / 10) % 10; 35 | //相加 36 | return num3 + num2 * 10 + num1 * 100; 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/reverse-bits.md: -------------------------------------------------------------------------------- 1 | # 颠倒二进制位 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reverse-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 颠倒给定的 32 位无符号整数的二进制位。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入: 00000010100101000001111010011100 9 | 输出: 00111001011110000010100101000000 10 | 解释: 11 | 输入的二进制串 00000010100101000001111010011100 表示无符号整数 43261596， 12 | 因此返回 964176192，其二进制表示形式为 00111001011110000010100101000000。 13 | ``` 14 | **样例 2:** 15 | ``` 16 | 输入：11111111111111111111111111111101 17 | 输出：10111111111111111111111111111111 18 | 解释： 19 | 输入的二进制串 11111111111111111111111111111101 表示无符号整数 4294967293， 20 | 因此返回 3221225471 其二进制表示形式为 10101111110010110010011101101001。 21 | ``` 22 | ### 参考代码 23 | 分离出输入数字的每一位，最低位左移31bit，次低位左移30bit，以此类推。 24 | ```java 25 | public class Solution { 26 | // you need treat n as an unsigned value 27 | public long reverseBits(int n) { 28 | int reversed = 0; 29 | for (int i = 0; i < 32; i++) { 30 | reversed = (reversed << 1) | (n & 1); 31 | n = (n >> 1); 32 | } 33 | return reversed; 34 | } 35 | } 36 | ``` 37 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/reverse-linked-list.md: -------------------------------------------------------------------------------- 1 | # 翻转链表 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reverse-linked-list/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

翻转一个链表

5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: 1->2->3->null 10 | 输出: 3->2->1->null 11 | ``` 12 | 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: 1->2->3->4->null 17 | 输出: 4->3->2->1->null 18 | ``` 19 | ### 参考代码 20 | http://www.lintcode.com/zh-cn/problem/reverse-linked-list/ 21 | ```python 22 | class Solution: 23 | 24 | def reverse(self, head): 25 | #curt表示前继节点 26 | curt = None 27 | while head != None: 28 | #temp记录下一个节点，head是当前节点 29 | temp = head.next 30 | head.next = curt 31 | curt = head 32 | head = temp 33 | return curt 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/search-in-rotated-sorted-array-ii.md: -------------------------------------------------------------------------------- 1 | # 搜索旋转排序数组 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/search-in-rotated-sorted-array-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

跟进“搜索旋转排序数组”，假如有重复元素又将如何？

是否会影响运行时间复杂度？

如何影响？

为何会影响？

写出一个函数判断给定的目标值是否出现在数组中。

5 | ### 样例说明 6 | 例1: 7 | ``` 8 | 输入: 9 | [] 10 | 1 11 | 输出: 12 | false 13 | ``` 14 | 例2: 15 | ``` 16 | 输入: 17 | [3,4,4,5,7,0,1,2] 18 | 4 19 | 输出: 20 | true 21 | ``` 22 | 23 | ### 参考代码 24 | 由于出现重复元素，故不能使用二分法。 25 | 所以暴力比较即可。 26 | ```java 27 | public class Solution { 28 | // 这个问题在面试中不会让实现完整程序 29 | // 只需要举出能够最坏情况的数据是 [1,1,1,1... 1] 里有一个0即可。 30 | // 在这种情况下是无法使用二分法的，复杂度是O(n) 31 | // 因此写个for循环最坏也是O(n)，那就写个for循环就好了 32 | // 如果你觉得，不是每个情况都是最坏情况，你想用二分法解决不是最坏情况的情况，那你就写一个二分吧。 33 | // 反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。 34 | public boolean search(int[] A, int target) { 35 | for (int i = 0; i < A.length; i ++) { 36 | if (A[i] == target) { 37 | return true; 38 | } 39 | } 40 | return false; 41 | } 42 | } 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/second-max-of-array.md: -------------------------------------------------------------------------------- 1 | # 数组第二大数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/second-max-of-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在数组中找到第二大的数。 5 | ### 样例说明 6 | 例1： 7 | ``` 8 | 输入：[1,3,2,4] 9 | 输出：3 10 | ``` 11 | 例2： 12 | ``` 13 | 输入：[1,1,2,2] 14 | 输出：2 15 | ``` 16 | ### 参考代码 17 | 维护最大数和此大数即可 18 | ```java 19 | public class Solution { 20 | /** 21 | * @param nums: An integer array. 22 | * @return: The second max number in the array. 23 | */ 24 | public int secondMax(int[] nums) { 25 | int max = Math.max(nums[0], nums[1]); 26 | int second = Math.min(nums[0], nums[1]); 27 | 28 | for (int i = 2; i < nums.length; i++) { 29 | if (nums[i] > max) { 30 | second = max; 31 | max = nums[i]; 32 | } else if (nums[i] > second) { 33 | second = nums[i]; 34 | } 35 | } 36 | 37 | return second; 38 | } 39 | } 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/simplify-path.md: -------------------------------------------------------------------------------- 1 | # 简化路径 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/simplify-path/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个文件的绝对路径(Unix-style)，请进行路径简化。 5 | 6 | Unix中, `.` 表示当前目录, `..` 表示父目录。 7 | 8 | 结果必须以 `/` 开头，并且两个目录名之间有且只有一个 `/`。最后一个目录名(如果存在)后不能出现 `/` 。你需要保证结果是正确表示路径的最短的字符串。 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: "/home/" 14 | 输出: "/home" 15 | ``` 16 | 17 | **样例 2:** 18 | 19 | ``` 20 | 输入: "/a/./../../c/" 21 | 输出: "/c" 22 | 解释: "/" 没有上级目录, "/../" 的结果就是 "/". 23 | ``` 24 | ### 参考代码 25 | 用栈处理即可. 26 | 27 | 将原字符串以 `'/'` 分隔, 然后遍历: 28 | 29 | - 遇到正常的目录名, 入栈 30 | - 遇到 `'.'` 或 空名称 (对应 `"//"`) 则忽略 31 | - 遇到 `".."` 则从栈顶弹出一个元素 (如果栈为空则不弹栈, 对应 `"/../"`) 32 | 33 | 最后将栈中的元素以 `'/'` 连接得到结果. 34 | ```python 35 | class Solution: 36 | """ 37 | @param path: the original path 38 | @return: the simplified path 39 | """ 40 | def simplifyPath(self, path): 41 | path = path.split('/') 42 | stack = [] 43 | for i in path: 44 | if i == '..': 45 | if len(stack): 46 | stack.pop() 47 | elif i != '.' and i != '': 48 | stack.append(i) 49 | return '/' + '/'.join(stack) 50 | ``` 51 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/single-number.md: -------------------------------------------------------------------------------- 1 | # 落单的数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/single-number/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出 `2 * n + 1`个数字，除其中一个数字之外其他每个数字均出现两次，找到这个数字。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入：[1,1,2,2,3,4,4] 9 | 输出：3 10 | 解释： 11 | 仅3出现一次 12 | ``` 13 | **样例 2:** 14 | ``` 15 | 输入：[0,0,1] 16 | 输出：1 17 | 解释： 18 | 仅1出现一次 19 | ``` 20 | 21 | ### 参考代码 22 | Given an array of integers, every element appears twice except for one. Find that single one. 23 | 考点： 24 | * 异或位运算 25 | 26 | Note: 27 | Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 28 | 异或运算具有很好的性质，相同数字异或运算后为0，并且具有交换律和结合律，故将所有数字异或运算后即可得到只出现一次的数字。 29 | ```python 30 | class Solution: 31 | """ 32 | @param A : an integer array 33 | @return : a integer 34 | """ 35 | def singleNumber(self, A): 36 | # write your code here 37 | ans = 0; 38 | for x in A: 39 | ans = ans ^ x 40 | return ans 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/singleton.md: -------------------------------------------------------------------------------- 1 | # 单例 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/singleton/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | **单例** 是最为最常见的设计模式之一。对于任何时刻，如果某个类只存在且最多存在一个具体的实例，那么我们称这种设计模式为单例。例如，对于 class Mouse (不是动物的mouse哦)，我们应将其设计为 singleton 模式。 5 | 6 | 你的任务是设计一个 `getInstance` 方法，对于给定的类，每次调用 `getInstance` 时，都可得到同一个实例。 7 | ### 样例说明 8 | ``` 9 | 在 Java 中: 10 | 11 | A a = A.getInstance(); 12 | A b = A.getInstance(); 13 | 14 | a 应等于 b. 15 | 16 | ``` 17 | ### 参考代码 18 | 创建私有静态成员对象，每次获取时返回此对象。 19 | ```java 20 | class Solution { 21 | 22 | 23 | /** 24 | * @return: The same instance of this class every time 25 | */ 26 | public static Solution instance = null; 27 | public static Solution getInstance() { 28 | if (instance == null) { 29 | instance = new Solution(); 30 | } 31 | return instance; 32 | } 33 | }; 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/sort-letters-by-case.md: -------------------------------------------------------------------------------- 1 | # 字符大小写排序 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/sort-letters-by-case/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给定一个只包含字母的字符串，按照先小写字母后大写字母的顺序进行排序。

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: "abAcD" 9 | 输出: "acbAD" 10 | 11 | 样例 2: 12 | 输入: "ABC" 13 | 输出: "ABC" 14 | 15 | ``` 16 | ### 参考代码 17 | ```java 18 | public class Solution { 19 | public void sortLetters(char[] chars) { 20 | int i = 0, j = chars.length - 1; 21 | 22 | while (i <= j) { 23 | while (i <= j && Character.isLowerCase(chars[i])) { 24 | i++; 25 | } 26 | while (i <= j && Character.isUpperCase(chars[j])) { 27 | j--; 28 | } 29 | if (i <= j) { 30 | char tmp = chars[i]; 31 | chars[i] = chars[j]; 32 | chars[j] = tmp; 33 | i++; 34 | j--; 35 | } 36 | } 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/sqrtx-ii.md: -------------------------------------------------------------------------------- 1 | # 对x开根II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/sqrtx-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 实现 `double sqrt(double x)` 并且 `x >= 0`。 5 | 计算并返回x开根后的值。 6 | ### 样例说明 7 | 例1: 8 | ``` 9 | 输入: n = 2 10 | 输出: 1.41421356 11 | ``` 12 | 例2: 13 | ``` 14 | 输入: n = 3 15 | 输出: 1.73205081 16 | ``` 17 | 18 | ### 参考代码 19 | ```python 20 | class Solution: 21 | """ 22 | @param: x: a double 23 | @return: the square root of x 24 | """ 25 | def sqrt(self, x): 26 | if x >= 1: 27 | start, end = 1, x 28 | else: 29 | start, end = x, 1 30 | 31 | while end - start > 1e-10: 32 | mid = (start + end) / 2 33 | if mid * mid < x: 34 | start = mid 35 | else: 36 | end = mid 37 | 38 | return start 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/string-permutation.md: -------------------------------------------------------------------------------- 1 | # 字符串置换 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/string-permutation/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定两个字符串，请设计一个方法来判定其中一个字符串是否为另一个字符串的置换。 5 | 6 | 置换的意思是，通过改变顺序可以使得两个字符串相等。 7 | ### 样例说明 8 | 9 | ### 参考代码 10 | 统计字符数量，组成成分一致即可置换得到。 11 | ```java 12 | public class Solution { 13 | /** 14 | * @param A a string 15 | * @param B a string 16 | * @return a boolean 17 | */ 18 | public boolean stringPermutation(String A, String B) { 19 | // Write your code here 20 | int[] cnt = new int[1000]; 21 | for (int i = 0; i < A.length(); ++i) 22 | cnt[(int)A.charAt(i)] += 1; 23 | for (int i = 0; i < B.length(); ++i) 24 | cnt[(int)B.charAt(i)] -= 1; 25 | for (int i = 0; i < 1000; ++i) 26 | if (cnt[i] != 0) 27 | return false; 28 | return true; 29 | } 30 | } 31 | ``` 32 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/subarray-sum.md: -------------------------------------------------------------------------------- 1 | # 子数组之和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/subarray-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组，找到和为 $0$ 的子数组。你的代码应该返回满足要求的子数组的起始位置和结束位置 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: [-3, 1, 2, -3, 4] 10 | 输出: [0,2] 或 [1,3] 11 | 样例解释： 返回任意一段和为0的区间即可。 12 | ``` 13 | 14 | **样例 2:** 15 | 16 | ``` 17 | 输入: [-3, 1, -4, 2, -3, 4] 18 | 输出: [1,5] 19 | ``` 20 | ### 参考代码 21 | 用九章算法班里讲过的前缀和 22 | 23 | 某一段[l, r]的和为0， 则其对应presum[l-1] = presum[r]. 24 | presum 为数组前缀和。只要保存每个前缀和，找是否有相同的前缀和即可。 25 | ```python 26 | class Solution: 27 | """ 28 | @param nums: A list of integers 29 | @return: A list of integers includes the index of the first number and the index of the last number 30 | """ 31 | def subarraySum(self, nums): 32 | prefix_hash = {0: -1} 33 | prefix_sum = 0 34 | for i, num in enumerate(nums): 35 | prefix_sum += num 36 | if prefix_sum in prefix_hash: 37 | return prefix_hash[prefix_sum] + 1, i 38 | prefix_hash[prefix_sum] = i 39 | 40 | return -1, -1 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/subsets-ii.md: -------------------------------------------------------------------------------- 1 | # 子集 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/subsets-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个可能具有重复数字的列表，返回其所有可能的子集。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：[0] 9 | 输出： 10 | [ 11 | [], 12 | [0] 13 | ] 14 | ``` 15 | **样例 2：** 16 | ``` 17 | 输入：[1,2,2] 18 | 输出： 19 | [ 20 | [2], 21 | [1], 22 | [1,2,2], 23 | [2,2], 24 | [1,2], 25 | [] 26 | ] 27 | ``` 28 | ### 参考代码 29 | 九章算法班中的方法 30 | ```python 31 | class Solution: 32 | """ 33 | @param nums: A set of numbers. 34 | @return: A list of lists. All valid subsets. 35 | """ 36 | def subsetsWithDup(self, nums): 37 | nums = sorted(nums) 38 | subsets = [] 39 | self.dfs(nums, 0, [], subsets) 40 | return subsets 41 | 42 | def dfs(self, nums, index, subset, subsets): 43 | subsets.append(list(subset)) 44 | 45 | for i in range(index, len(nums)): 46 | if i > index and nums[i] == nums[i - 1]: 47 | continue 48 | subset.append(nums[i]) 49 | self.dfs(nums, i + 1, subset, subsets) 50 | subset.pop() 51 | ``` 52 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/swap-bits.md: -------------------------------------------------------------------------------- 1 | # 交换奇偶二进制位 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/swap-bits/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 设计一个方法，用尽可能少的指令，将一个整数中奇数数位和偶数数位的数字交换 （如，数位 0 和数位 1 交换，数位 2 和数位 3 交换，等等）。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：0 9 | 输出：0 10 | 解释： 11 | 0 = 0(2) -> 0(2) = 0 12 | ``` 13 | **样例 2：** 14 | ``` 15 | 输入：5 16 | 输出：10 17 | 解释： 18 | 5 = 101(2) -> 1010(2) = 10 19 | ``` 20 | ### 参考代码 21 | ```cpp 22 | class Solution { 23 | public: 24 | /** 25 | * @param x an integer 26 | * @return an integer 27 | */ 28 | int swapOddEvenBits(int x) { 29 | // Write your code here 30 | return ( ((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1) ); 31 | } 32 | }; 33 | ``` 34 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/swap-two-integers-in-array.md: -------------------------------------------------------------------------------- 1 | # 交换数组两个元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/swap-two-integers-in-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你一个数组和两个索引，交换下标为这两个索引的数字 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: [1, 2, 3, 4], index1 = 2, index2 = 3 10 | 输出: 交换后你的数组应该是[1, 2, 4, 3]， 不需要返回任何值，只要就地对数组进行交换即可。 11 | 样例解释: 就地交换，不需要返回值。 12 | ``` 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: [1, 2, 2, 2], index1 = 0, index2 = 3 17 | 输出: 交换后你的数组应该是[2, 2, 2, 1]， 不需要返回任何值，只要就地对数组进行交换即可。 18 | 样例解释: 就地交换，不需要返回值。 19 | ``` 20 | ### 参考代码 21 | 模拟即可 22 | ```java 23 | public class Solution { 24 | /** 25 | * @param A an integer array 26 | * @param index1 the first index 27 | * @param index2 the second index 28 | * @return void 29 | */ 30 | public void swapIntegers(int[] A, int index1, int index2) { 31 | // Write your code here 32 | int tmp = A[index1]; 33 | A[index1] = A[index2]; 34 | A[index2] = tmp; 35 | } 36 | } 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/swap-without-extra-variable-only-c.md: -------------------------------------------------------------------------------- 1 | # 交换不借助额外变量(仅C++) 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/swap-without-extra-variable-only-c/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出两个变量，x 和 y ，不使用第三个变量交换这两个变量的值。 5 | ### 样例说明 6 | **样例 1:** 7 | ``` 8 | 输入 : x = 10, y = 5 9 | 输出 : x = 5, y = 10 10 | ``` 11 | ### 参考代码 12 | 通过位运算实现。 13 | ```cpp 14 | class Solution { 15 | public: 16 | /** 17 | * @param x an integer 18 | * @param y an integer 19 | * @return nothing 20 | */ 21 | void swap(int &x, int &y) { 22 | // Write your code here 23 | x = x ^ y; 24 | y = x ^ y; 25 | x = x ^ y; 26 | } 27 | }; 28 | ``` 29 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/trailing-zeros.md: -------------------------------------------------------------------------------- 1 | # 尾部的零 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/trailing-zeros/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

设计一个算法，计算出n阶乘中尾部零的个数

5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 11 9 | 输出: 2 10 | 11 | 样例解释: 12 | 11! = 39916800, 结尾的0有2个。 13 | 14 | 样例 2: 15 | 输入: 5 16 | 输出: 1 17 | 18 | 样例解释: 19 | 5! = 120， 结尾的0有1个。 20 | 21 | ``` 22 | ### 参考代码 23 | 可以将每个数拆分成其素因子的乘积，可以发现，0是由`2*5`产生的，而5的数量一定小于2的数量，因此5的个数决定了结尾0的个数。 24 | 25 | 只要计算n的阶乘中，5这个素因子出现多少次即可。 26 | ```java 27 | class Solution { 28 | /* 29 | * param n: As desciption 30 | * return: An integer, denote the number of trailing zeros in n! 31 | */ 32 | public long trailingZeros(long n) { 33 | long sum = 0; 34 | while (n != 0) { 35 | sum += n / 5; 36 | n /= 5; 37 | } 38 | return sum; 39 | } 40 | }; 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/ugly-number-ii.md: -------------------------------------------------------------------------------- 1 | # 丑数 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/ugly-number-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 设计一个算法，找出只含素因子`2`，`3`，`5` 的第 *n* 小的数。 5 | 6 | 符合条件的数如：`1, 2, 3, 4, 5, 6, 8, 9, 10, 12...` 7 | 8 | ### 样例说明 9 | **样例 1：** 10 | ``` 11 | 输入：9 12 | 输出：10 13 | ``` 14 | **样例 2：** 15 | ``` 16 | 输入：1 17 | 输出：1 18 | ``` 19 | ### 参考代码 20 | 使用九章算法班中讲的 heap 21 | 时间复杂度 $O(nlogn)$ 22 | ```python 23 | import heapq 24 | 25 | class Solution: 26 | """ 27 | @param {int} n an integer. 28 | @return {int} the nth prime number as description. 29 | """ 30 | def nthUglyNumber(self, n): 31 | heap = [1] 32 | visited = set([1]) 33 | 34 | val = None 35 | for i in range(n): 36 | val = heapq.heappop(heap) 37 | for factor in [2, 3, 5]: 38 | if val * factor not in visited: 39 | visited.add(val * factor) 40 | heapq.heappush(heap, val * factor) 41 | 42 | return val 43 | ``` 44 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/ugly-number.md: -------------------------------------------------------------------------------- 1 | # 丑数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/ugly-number/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 写一个程序来检测一个整数是不是`丑数`。 5 | 6 | 丑数的定义是，只包含质因子 `2, 3, 5` 的正整数。比如 6, 8 就是丑数，但是 14 不是丑数因为他包含了质因子 7。 7 | ### 样例说明 8 | 例1: 9 | ``` 10 | 输入: num = 8 11 | 输出: true 12 | 解释: 13 | 8=2*2*2 14 | ``` 15 | 例2: 16 | ``` 17 | 输入: num = 14 18 | 输出: false 19 | 解释: 20 | 14=2*7 21 | ``` 22 | 23 | ### 参考代码 24 | 看这个数是否只有2，3，5的因子 25 | ```java 26 | public class Solution { 27 | /** 28 | * @param num an integer 29 | * @return true if num is an ugly number or false 30 | */ 31 | public boolean isUgly(int num) { 32 | if (num <= 0) return false; 33 | if (num == 1) return true; 34 | 35 | while (num >= 2 && num % 2 == 0) num /= 2; 36 | while (num >= 3 && num % 3 == 0) num /= 3; 37 | while (num >= 5 && num % 5 == 0) num /= 5; 38 | 39 | return num == 1; 40 | } 41 | } 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/unique-characters.md: -------------------------------------------------------------------------------- 1 | # 判断字符串是否没有重复字符 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/unique-characters/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 实现一个算法确定字符串中的字符是否均唯一出现 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: "abc_____" 10 | 输出: false 11 | ``` 12 | 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: "abc" 17 | 输出: true 18 | ``` 19 | ### 参考代码 20 | 使用一个map或数组标记每个字符是否出现过。 21 | ```java 22 | public class Solution { 23 | /** 24 | * @param str: a string 25 | * @return: a boolean 26 | */ 27 | public boolean isUnique(String str) { 28 | // write your code here 29 | boolean[] char_set = new boolean[256]; 30 | for (int i = 0; i < str.length(); i++) { 31 | int val = str.charAt(i); 32 | if (char_set[val]) return false; 33 | char_set[val] = true; 34 | } 35 | return true; 36 | } 37 | } 38 | ``` 39 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /题解/window-sum.md: -------------------------------------------------------------------------------- 1 | # 滑动窗口内数的和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/window-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你一个大小为n的整型数组和一个大小为k的滑动窗口，将滑动窗口从头移到尾，输出从开始到结束每一个时刻滑动窗口内的数的和。 5 | ### 样例说明 6 | **样例 1** 7 | ``` 8 | 输入：array = [1,2,7,8,5], k = 3 9 | 输出：[10,17,20] 10 | 解析： 11 | 1 + 2 + 7 = 10 12 | 2 + 7 + 8 = 17 13 | 7 + 8 + 5 = 20 14 | ``` 15 | ### 参考代码 16 | ```python 17 | class Solution: 18 | # @param nums {int[]} a list of integers 19 | # @param k {int} size of window 20 | # @return {int[]} the sum of element inside the window at each moving 21 | def winSum(self, nums, k): 22 | # Write your code here 23 | n = len(nums) 24 | if n < k or k <= 0: 25 | return [] 26 | sums = [0] * (n - k + 1) 27 | for i in range(k): 28 | sums[0] += nums[i]; 29 | 30 | for i in range(1, n - k + 1): 31 | sums[i] = sums[i - 1] - nums[i - 1] + nums[i + k - 1] 32 | 33 | return sums 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/anagrams.md: -------------------------------------------------------------------------------- 1 | # 乱序字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/anagrams/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 |

给出一个字符串数组S，找到其中所有的乱序字符串(Anagram)。如果一个字符串是乱序字符串，那么他存在一个字母集合相同，但顺序不同的字符串也在S中。

5 | ### 样例说明 6 | **样例1:** 7 | ``` 8 | 输入:["lint", "intl", "inlt", "code"] 9 | 输出:["lint", "inlt", "intl"] 10 | ``` 11 | 12 | **样例 2:** 13 | ``` 14 | 输入:["ab", "ba", "cd", "dc", "e"] 15 | 输出: ["ab", "ba", "cd", "dc"] 16 | ``` 17 | ### 参考代码 18 | Given an array of strings, return all groups of strings that are anagrams. 19 | 20 | Note: All inputs will be in lower-case 21 | ```python 22 | class Solution: 23 | # @param strs: A list of strings 24 | # @return: A list of strings 25 | def anagrams(self, strs): 26 | # write your code here 27 | dict = {} 28 | for word in strs: 29 | sortedword = ''.join(sorted(word)) 30 | dict[sortedword] = [word] if sortedword not in dict else dict[sortedword] + [word] 31 | res = [] 32 | for item in dict: 33 | if len(dict[item]) >= 2: 34 | res += dict[item] 35 | return res 36 | 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/char-to-integer.md: -------------------------------------------------------------------------------- 1 | # 字符转整数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/char-to-integer/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将字符转换为一个整数。你可以假设字符是ASCII码，也就是说转换后的整数在0~255之间。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: 'a' 9 | 输出: 97 10 | 11 | 样例解释: 12 | 返回ASCII码中对应的数字. 13 | 14 | 样例 2: 15 | 输入: '%' 16 | 输出: 37 17 | 18 | 样例解释: 19 | 返回ASCII码中对应的数字. 20 | 21 | ``` 22 | ### 参考代码 23 | 类型转换 24 | ```java 25 | public class Solution { 26 | /** 27 | * @param character a character 28 | * @return an integer 29 | */ 30 | public int charToInteger(char character) { 31 | return (int) character; 32 | } 33 | } 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/compare-strings.md: -------------------------------------------------------------------------------- 1 | # 比较字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/compare-strings/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 比较两个字符串A和B，确定A中是否包含B中所有的字符。字符串A和B中的字符都是 **大写字母** 5 | ### 样例说明 6 | 给出 A = `"ABCD"` B = `"ACD"`，返回 `true` 7 | 8 | 给出 A = `"ABCD"` B = `"AABC"`， 返回 `false` 9 | ### 参考代码 10 | ```python 11 | class Solution: 12 | """ 13 | @param A : A string includes Upper Case letters 14 | @param B : A string includes Upper Case letters 15 | @return : if string A contains all of the characters in B return True else return False 16 | """ 17 | def compareStrings(self, A, B): 18 | # write your code here 19 | if len(B) == 0: 20 | return True 21 | if len(A) == 0: 22 | return False 23 | #trackTable首先记录A中所有的字符以及它们的个数，然后遍历B,如果出现trackTable[i]小于0的情况，说明B中该字符出现的次数 24 | #大于在A中出现的次数 25 | trackTable = [0 for _ in range(26)] 26 | for i in A: 27 | trackTable[ord(i) - 65] += 1 28 | for i in B: 29 | if trackTable[ord(i) - 65] == 0: 30 | return False 31 | else: 32 | trackTable[ord(i) -65] -= 1 33 | return True 34 | ``` 35 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/excel-sheet-column-title.md: -------------------------------------------------------------------------------- 1 | # Excel表列标题 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/excel-sheet-column-title/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个正整数，返回相应的列标题，如Excel表中所示。 5 | ### 样例说明 6 | **样例1** 7 | 8 | ``` 9 | 输入: 28 10 | 输出: "AB" 11 | ``` 12 | 13 | **样例2** 14 | 15 | ``` 16 | 输入: 29 17 | 输出: "AC" 18 | ``` 19 | ### 参考代码 20 | ```cpp 21 | class Solution { 22 | public: 23 | string convertToTitle(int n) { 24 | if (n == 0) { 25 | return ""; 26 | } 27 | return convertToTitle((n - 1) / 26) + (char)((n - 1) % 26 + 'A'); 28 | } 29 | }; 30 | ``` 31 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/first-unique-character-in-a-string.md: -------------------------------------------------------------------------------- 1 | # 第一个只出现一次的字符 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/first-unique-character-in-a-string/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出一个字符串，找出第一个只出现一次的字符。 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: "abaccdeff" 9 | 输出: 'b' 10 | 11 | 解释: 12 | 'b' 是第一个出现一次的字符 13 | 14 | 15 | 样例 2: 16 | 输入: "aabccd" 17 | 输出: 'b' 18 | 19 | 解释: 20 | 'b' 是第一个出现一次的字符 21 | 22 | 23 | ``` 24 | ### 参考代码 25 | 辅助数组记录字符出现次数。 26 | ```python 27 | class Solution: 28 | """ 29 | @param str: str: the given string 30 | @return: char: the first unique character in a given string 31 | """ 32 | def firstUniqChar(self, str): 33 | counter = {} 34 | 35 | for c in str: 36 | counter[c] = counter.get(c, 0) + 1 37 | 38 | for c in str: 39 | if counter[c] == 1: 40 | return c 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/is-subsequence.md: -------------------------------------------------------------------------------- 1 | # 是子序列吗？ 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/is-subsequence/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定字符串`s`和`t`，判断`s`是否为`t`的子序列。 5 | 6 | 你可以认为在`s`和`t`中都只包含小写字母。`t`可能是一个非常长（`length ~= 500,000`）的字符串，而`s`是一个较短的字符串（`length <= 100`）。 7 | 8 | 一个字符串的子序列是在原字符串中删去一些字符（也可以不删除）后，不改变剩余字符的相对位置形成的新字符串（例如，`"ace"`是`"abcde"`的子序列而`"aec"`不是）。 9 | ### 样例说明 10 | **样例1：** 11 | ``` 12 | 输入：s = "abc"，t = "ahbgdc" 13 | 输出：true 14 | ``` 15 | **样例2：** 16 | ``` 17 | 输入：s = "axc"，t = "ahbgdc" 18 | 输出：false 19 | ``` 20 | ### 参考代码 21 | 直接按s串的顺序去遍历t串即可。 22 | ```java 23 | public class Solution { 24 | public boolean isSubsequence(String s, String t) { 25 | int i = 0, j = 0; 26 | while(i 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/longest-words/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一个词典，找出其中所有最长的单词。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | ```java 9 | class Solution { 10 | /** 11 | * @param dictionary: an array of strings 12 | * @return: an arraylist of strings 13 | */ 14 | ArrayList longestWords(String[] dictionary) { 15 | // write your code here 16 | int maxLen = 0; 17 | ArrayList ans = new ArrayList(); 18 | for (int i=0; imaxLen) maxLen = dictionary[i].length(); 20 | for (int i=0; i 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/lowercase-to-uppercase-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个字符串中的小写字母转换为大写字母。不是字母的字符不需要做改变。 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: str = "abc" 10 | 输出: "ABC" 11 | ``` 12 | **样例 2:** 13 | 14 | ``` 15 | 输入: str = "aBc" 16 | 输出: "ABC" 17 | ``` 18 | **样例 3:** 19 | 20 | ``` 21 | 输入: str = "abC12" 22 | 输出: "ABC12" 23 | ``` 24 | ### 参考代码 25 | ```java 26 | public class Solution { 27 | /** 28 | * @param str a string 29 | * @return a string 30 | */ 31 | public String lowercaseToUppercase2(String str) { 32 | // Write your code here 33 | StringBuilder sb = new StringBuilder(str); 34 | //遍历整个字符串，将所有的小写字母转成大写字母 35 | for (int index = 0; index < sb.length(); index++) { 36 | char c = sb.charAt(index); 37 | if (Character.isLowerCase(c)) { 38 | sb.setCharAt(index, Character.toUpperCase(c)); 39 | } 40 | } 41 | return sb.toString(); 42 | } 43 | } 44 | ``` 45 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/lowercase-to-uppercase.md: -------------------------------------------------------------------------------- 1 | # 大小写转换 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/lowercase-to-uppercase/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个字符由小写字母转换为大写字母 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: 'a' 10 | 输出: 'A' 11 | ``` 12 | **样例 2:** 13 | 14 | ``` 15 | 输入: 'b' 16 | 输出: 'B' 17 | ``` 18 | ### 参考代码 19 | ```java 20 | public class Solution { 21 | /** 22 | * @param character a character 23 | * @return a character 24 | */ 25 | public char lowercaseToUppercase(char character) { 26 | // Write your code here 27 | //获得character与'a'的差值，在'A'的基础上加上这个差值 28 | return (char)(character - 'a' + 'A'); 29 | } 30 | } 31 | ``` 32 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/null-terminated-string.md: -------------------------------------------------------------------------------- 1 | # NULL 结尾的字符串 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/null-terminated-string/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 用 C 或 C++ 设计一个 void reverse `(char* str)` 功能，实现把一个以 NULL 结尾字符串翻转的功能。 5 | ### 样例说明 6 | 7 | ### 参考代码 8 | 首尾双指针，依次交换，直至相遇。 9 | ```cpp 10 | class Solution { 11 | public: 12 | /** 13 | * @param str C-String 14 | * @return void 15 | */ 16 | void reverse(char *str) { 17 | // Write your code here 18 | char * end = str; 19 | char tmp; 20 | if (str) { 21 | while (*end != NULL) { 22 | ++end; 23 | } 24 | } 25 | --end; 26 | while (str < end) { 27 | tmp = *str; 28 | *str++ = *end; 29 | *end-- = tmp; 30 | } 31 | } 32 | }; 33 | ``` 34 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/pass-interview.md: -------------------------------------------------------------------------------- 1 | # 通过面试 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/pass-interview/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给出面试者的性别和面试成绩，判断求职者是否通过面试。 5 | 6 | * 如果面试者是男性，那么面试成绩必须大于等于4分。 7 | * 如果面试者是女性，那么面试成绩必须大于等于3分。 8 | 9 | 参数sex是boolean类型，sex为true时，代表求职者是女性，因为女人永远都是对的。相反sex为false时，代表求职者为男性，因为男人永远都是错的。 10 | ### 样例说明 11 | 12 | ### 参考代码 13 | ```java 14 | public class Solution { 15 | /** 16 | * @param sex the sex of a candidate 17 | * @param score the interview score of a candidate 18 | * @return a boolean 19 | */ 20 | public boolean passInterview(boolean sex, double score) { 21 | if (sex) { 22 | return score >= 3; 23 | } 24 | return score >= 4; 25 | } 26 | } 27 | 28 | ``` 29 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/prime-factorization.md: -------------------------------------------------------------------------------- 1 | # 分解质因数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/prime-factorization/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 将一个整数分解为若干质因数之乘积。 5 | ### 样例说明 6 | **样例 1：** 7 | ``` 8 | 输入：10 9 | 输出：[2, 5] 10 | ``` 11 | **样例 2：** 12 | ``` 13 | 输入：660 14 | 输出：[2, 2, 3, 5, 11] 15 | ``` 16 | ### 参考代码 17 | ```java 18 | public class Solution { 19 | /** 20 | * @param num an integer 21 | * @return an integer array 22 | */ 23 | public List primeFactorization(int num) { 24 | List factors = new ArrayList(); 25 | 26 | for (int i = 2; i * i <= num; i++) { 27 | while (num % i == 0) { 28 | num = num / i; 29 | factors.add(i); 30 | } 31 | } 32 | 33 | if (num != 1) { 34 | factors.add(num); 35 | } 36 | 37 | return factors; 38 | } 39 | } 40 | 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/readme.md: -------------------------------------------------------------------------------- 1 | 2 | -------------------------------------------------------------------------------- /高频/reference.md: -------------------------------------------------------------------------------- 1 | # 引用 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/reference/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 实现一个类 *ReferenceManager* 包含如下两种方法 5 | 6 | 1.`copyValue(Node obj)` 只拷贝参数*obj*的权值，*obj*和*node*仍然是两个指针 7 | 8 | 2.`copyReference(Node obj)` *obj*和*node*指向同一个地方 9 | ### 样例说明 10 | ``` 11 | ReferenceManager ref = ReferenceManager(); 12 | Node obj = new Node(0); 13 | ref.copyValue(obj); 14 | ref.node.val; // 值为0 15 | ref.node; // 与obj不同. 16 | 17 | Node obj2 = new Node(1); 18 | ref.copyReference(obj2); 19 | ref.node.val; // 值为1 20 | ref.node; // 与 obj2相同，指向同一个地址 21 | ``` 22 | ### 参考代码 23 | java中直接赋值是同一个变量，指向同一个地址 24 | ```java 25 | public class ReferenceManager { 26 | public Node node; 27 | 28 | public void copyValue(Node obj) { 29 | if (obj == null) { 30 | return; 31 | } 32 | if (node == null) { 33 | node = new Node(obj.val); 34 | } 35 | node.val = obj.val; 36 | } 37 | 38 | public void copyReference(Node obj) { 39 | node = obj; 40 | } 41 | } 42 | ``` 43 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/best-time-to-buy-and-sell-stock-ii.md: -------------------------------------------------------------------------------- 1 | # 买卖股票的最佳时机 II 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/best-time-to-buy-and-sell-stock-ii/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个数组 `prices` 表示一支股票每天的价格. 5 | 6 | 你可以完成任意次数的交易, 不过你不能同时参与多个交易 (也就是说, 如果你已经持有这支股票, 在再次购买之前, 你必须先卖掉它). 7 | 8 | 设计一个算法求出最大的利润. 9 | ### 样例说明 10 | **样例 1:** 11 | 12 | ``` 13 | 输入: [2, 1, 2, 0, 1] 14 | 输出: 2 15 | 解释: 16 | 1. 在第 2 天以 1 的价格买入, 然后在第 3 天以 2 的价格卖出, 利润 1 17 | 2. 在第 4 天以 0 的价格买入, 然后在第 5 天以 1 的价格卖出, 利润 1 18 | 总利润 2. 19 | ``` 20 | 21 | **样例 2:** 22 | 23 | ``` 24 | 输入: [4, 3, 2, 1] 25 | 输出: 0 26 | 解释: 不进行任何交易, 利润为0. 27 | ``` 28 | ### 参考代码 29 | 贪心法: 只要相邻的两天股票的价格是上升的, 我们就进行一次交易, 获得一定利润. 30 | 31 | 这样的策略不一定是最小的交易次数, 但是一定会得到最大的利润. 32 | ```java 33 | public class Solution { 34 | public int maxProfit(int[] prices) { 35 | int profit = 0; 36 | for (int i = 0; i < prices.length - 1; i++) { 37 | int diff = prices[i + 1] - prices[i]; 38 | if (diff > 0) { 39 | profit += diff; 40 | } 41 | } 42 | return profit; 43 | } 44 | } 45 | ``` 46 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/can-place-flowers.md: -------------------------------------------------------------------------------- 1 | # 能否放置花 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/can-place-flowers/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 假设你有一个长花圃，其中有些地块已经被种植了，但是有些地块没有。但是，花不能够在相邻的地块下种植 - 他们会争夺水从而导致两者的死亡。 5 | 6 | 给定一个花圃（用一个包含0和1的数组来表示，其中0代表空，1代表非空），和一个数字**n**，返回n朵新的花在这个花圃上以能否在不违反“无相邻花”的规则种植。 7 | ### 样例说明 8 | **样例1** 9 | ``` 10 | 输入: flowerbed = [1,0,0,0,1], n = 1 11 | 输出: True 12 | ``` 13 | **样例2** 14 | ``` 15 | 输入: flowerbed = [1,0,0,0,1], n = 2 16 | 输出: False 17 | ``` 18 | ### 参考代码 19 | 比较基础的模拟题，按照题目要求从头开始模拟种植即可。 20 | ```java 21 | public class Solution { 22 | public boolean canPlaceFlowers(int[] flowerbed, int n) { 23 | int count = 0; 24 | for (int i = 0; i < flowerbed.length; i++) { 25 | if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] == 0) 26 | && (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) { 27 | flowerbed[i] = 1; 28 | count++; 29 | } 30 | if (count >= n) { 31 | return true; 32 | } 33 | } 34 | return false; 35 | } 36 | } 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/generate-arraylist-with-given-size.md: -------------------------------------------------------------------------------- 1 | # 生成给定大小的数组 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/generate-arraylist-with-given-size/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你一个大小size,生成一个元素`从1 到 size`的数组 5 | ### 样例说明 6 | ``` 7 | 样例 1: 8 | 输入: size = 4 9 | 输出: [1, 2, 3, 4] 10 | 11 | 样例解释: 12 | 返回一个顺序填充1到4的数组。 13 | 14 | 样例 2: 15 | 输入: size = 1 16 | 输出: [1] 17 | 18 | 样例解释: 19 | 返回一个顺序填充1到1的数组 20 | 21 | ``` 22 | 23 | ### 参考代码 24 | 基本模拟 25 | ```java 26 | public class Solution { 27 | /** 28 | * @param size an integer 29 | * @return an array list 30 | */ 31 | public ArrayList generate(int size) { 32 | // Write your code here 33 | ArrayList result = new ArrayList(); 34 | for (int i = 1; i <= size; ++i) 35 | result.add(i); 36 | return result; 37 | } 38 | } 39 | ``` 40 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/max-of-array.md: -------------------------------------------------------------------------------- 1 | # 数组的最大值 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/max-of-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给一个浮点数数组，求数组中的最大值。 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: [1.0, 2.1, -3.3] 10 | 输出: 2.1 11 | 样例解释: 返回最大的数字 12 | ``` 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: [1.0, 1.0, -3.3] 17 | 输出: 1.0 18 | 样例解释: 返回最大的数字。 19 | ``` 20 | ### 参考代码 21 | 模拟过程 22 | ```java 23 | public class Solution { 24 | /** 25 | * @param A a float array 26 | * @return a float number 27 | */ 28 | public float maxOfArray(float[] A) { 29 | float max = A[0]; 30 | for (int i = 1; i < A.length; i++) { 31 | if (A[i] > max) { 32 | max = A[i]; 33 | } 34 | } 35 | return max; 36 | } 37 | } 38 | ``` 39 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/minimum-moves-to-equal-array-elements.md: -------------------------------------------------------------------------------- 1 | # 使数组元素相同的最少步数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/minimum-moves-to-equal-array-elements/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个大小为`n`的**非空**整数数组，找出使得数组中所有元素相同的最少步数，其中一步被定义为将数组中`n - 1`个元素加一。 5 | ### 样例说明 6 | ``` 7 | 输入： 8 | [1,2,3] 9 | 10 | 输出： 11 | 3 12 | 13 | 说明： 14 | 只需要三步即可（每一步将其中两个元素加一）： 15 | 16 | [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4] 17 | ``` 18 | ### 参考代码 19 | ```cpp 20 | class Solution { 21 | public: 22 | int minMoves(vector& nums) { 23 | int sumNum = 0; 24 | int minNum = INT_MAX; 25 | for (int num : nums) { 26 | sumNum += num; 27 | minNum = min(minNum, num); 28 | } 29 | return sumNum - minNum * nums.size(); 30 | } 31 | }; 32 | ``` 33 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/second-max-of-array.md: -------------------------------------------------------------------------------- 1 | # 数组第二大数 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/second-max-of-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 在数组中找到第二大的数。 5 | ### 样例说明 6 | 例1： 7 | ``` 8 | 输入：[1,3,2,4] 9 | 输出：3 10 | ``` 11 | 例2： 12 | ``` 13 | 输入：[1,1,2,2] 14 | 输出：2 15 | ``` 16 | ### 参考代码 17 | 维护最大数和此大数即可 18 | ```java 19 | public class Solution { 20 | /** 21 | * @param nums: An integer array. 22 | * @return: The second max number in the array. 23 | */ 24 | public int secondMax(int[] nums) { 25 | int max = Math.max(nums[0], nums[1]); 26 | int second = Math.min(nums[0], nums[1]); 27 | 28 | for (int i = 2; i < nums.length; i++) { 29 | if (nums[i] > max) { 30 | second = max; 31 | max = nums[i]; 32 | } else if (nums[i] > second) { 33 | second = nums[i]; 34 | } 35 | } 36 | 37 | return second; 38 | } 39 | } 40 | ``` 41 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/subarray-sum.md: -------------------------------------------------------------------------------- 1 | # 子数组之和 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/subarray-sum/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给定一个整数数组，找到和为 $0$ 的子数组。你的代码应该返回满足要求的子数组的起始位置和结束位置 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: [-3, 1, 2, -3, 4] 10 | 输出: [0,2] 或 [1,3] 11 | 样例解释： 返回任意一段和为0的区间即可。 12 | ``` 13 | 14 | **样例 2:** 15 | 16 | ``` 17 | 输入: [-3, 1, -4, 2, -3, 4] 18 | 输出: [1,5] 19 | ``` 20 | ### 参考代码 21 | 用九章算法班里讲过的前缀和 22 | 23 | 某一段[l, r]的和为0， 则其对应presum[l-1] = presum[r]. 24 | presum 为数组前缀和。只要保存每个前缀和，找是否有相同的前缀和即可。 25 | ```python 26 | class Solution: 27 | """ 28 | @param nums: A list of integers 29 | @return: A list of integers includes the index of the first number and the index of the last number 30 | """ 31 | def subarraySum(self, nums): 32 | prefix_hash = {0: -1} 33 | prefix_sum = 0 34 | for i, num in enumerate(nums): 35 | prefix_sum += num 36 | if prefix_sum in prefix_hash: 37 | return prefix_hash[prefix_sum] + 1, i 38 | prefix_hash[prefix_sum] = i 39 | 40 | return -1, -1 41 | ``` 42 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/数组题/swap-two-integers-in-array.md: -------------------------------------------------------------------------------- 1 | # 交换数组两个元素 2 | > 题目来源: [LintCode 算法题库](https://www.lintcode.com/problem/swap-two-integers-in-array/?utm_source=sc-github-wzz) 3 | ## 题目描述 4 | 给你一个数组和两个索引，交换下标为这两个索引的数字 5 | ### 样例说明 6 | **样例 1:** 7 | 8 | ``` 9 | 输入: [1, 2, 3, 4], index1 = 2, index2 = 3 10 | 输出: 交换后你的数组应该是[1, 2, 4, 3]， 不需要返回任何值，只要就地对数组进行交换即可。 11 | 样例解释: 就地交换，不需要返回值。 12 | ``` 13 | **样例 2:** 14 | 15 | ``` 16 | 输入: [1, 2, 2, 2], index1 = 0, index2 = 3 17 | 输出: 交换后你的数组应该是[2, 2, 2, 1]， 不需要返回任何值，只要就地对数组进行交换即可。 18 | 样例解释: 就地交换，不需要返回值。 19 | ``` 20 | ### 参考代码 21 | 模拟即可 22 | ```java 23 | public class Solution { 24 | /** 25 | * @param A an integer array 26 | * @param index1 the first index 27 | * @param index2 the second index 28 | * @return void 29 | */ 30 | public void swapIntegers(int[] A, int index1, int index2) { 31 | // Write your code here 32 | int tmp = A[index1]; 33 | A[index1] = A[index2]; 34 | A[index2] = tmp; 35 | } 36 | } 37 | ``` 38 | 更多题解请参考: [LintCode 提供 2000+ 算法题和名企阶梯训练](https://www.lintcode.com/problem/?utm_source=sc-github-wzz) -------------------------------------------------------------------------------- /高频/算法面试必刷100题.xlsx: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/huangyingw/ninechapter-algorithm_linghu-algorithm-templete/ff7284350de9097925a41b59f293f1f1dda25622/高频/算法面试必刷100题.xlsx --------------------------------------------------------------------------------