├── .idea ├── misc.xml ├── modules.xml ├── uiDesigner.xml ├── vcs.xml └── workspace.xml ├── LeetCode.iml ├── README.md ├── images ├── _011_Container_With_Most_Water.jpg ├── _042_Trapping_Rain_Water.png ├── _043_Multiply_Strings.jpg ├── _309_Best_Time_to_Buy_and_Sell_Stock_with_Cooldown.png └── _714_Best_Time_to_Buy_and_Sell_Stock_with_Transaction_Fee.png ├── out └── production │ └── LeetCode │ └── algorithm │ ├── _001_Two_Sum │ └── Solution.class │ ├── _002_Add_Two_Numbers │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _003_Longest_Substring_Without_Repeating_Characters │ └── Solution.class │ ├── _004_Median_of_Two_Sorted_Arrays │ └── Solution.class │ ├── _005_Longest_Palindromic_Substring │ └── Solution.class │ ├── _007_Reverse_Integer │ └── Solution.class │ ├── _008_String_to_Integer │ └── Solution.class │ ├── _009_Palindrome_Number │ └── Solution.class │ ├── _011_Container_With_Most_Water │ └── Solution.class │ ├── _012_Integer_to_Roman │ └── Solution.class │ ├── _013_Roman_to_Integer │ └── Solution.class │ ├── _014_Longest_Common_Prefix │ └── Solution.class │ ├── _015_3Sum │ └── Solution.class │ ├── _016_3Sum_Closest │ └── Solution.class │ ├── _017_Letter_Combinations_of_a_Phone_Number │ └── Solution.class │ ├── _019_Remove_Nth_Node_From_End_of_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _020_Valid_Parentheses │ └── Solution.class │ ├── _021_Merge_Two_Sorted_Lists │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _022_Generate_Parentheses │ └── Solution.class │ ├── _023_Merge_k_Sorted_Lists │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _024_Swap_Nodes_in_Pairs │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _025_Reverse_Nodes_in_k_Group │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _027_Remove_Element │ └── Solution.class │ ├── _028_Implement_strStr │ └── Solution.class │ ├── _031_Next_Permutation │ └── Solution.class │ ├── _032_Longest_Valid_Parentheses │ └── Solution.class │ ├── _033_Search_in_Rotated_Sorted_Array │ └── Solution.class │ ├── _034_Find_First_and_Last_Position_of_Element_in_Sorted_Array │ └── Solution.class │ ├── _035_Search_Insert_Position │ └── Solution.class │ ├── _038_Count_and_Say │ └── Solution.class │ ├── _039_Combination_Sum │ └── Solution.class │ ├── _040_Combination_Sum_ll │ └── Solution.class │ ├── _041_First_Missing_Positive │ └── Solution.class │ ├── _042_Trapping_Rain_Water │ └── Solution.class │ ├── _043_Multiply_Strings │ └── Solution.class │ ├── _045_Jump_Game_II │ └── Solution.class │ ├── _046_Permutations │ ├── Solution.class │ └── Solution2.class │ ├── _047_Permutations_II │ └── Solution.class │ ├── _048_Rotate_Image │ └── Solution.class │ ├── _049_Group_Anagrams │ └── Solution.class │ ├── _050_Pow │ └── Solution.class │ ├── _053_Maximum_Subarray │ └── Solution.class │ ├── _054_Spiral_Matrix │ └── Solution.class │ ├── _055_Jump_Game │ └── Solution.class │ ├── _056_Merge_Intervals │ └── Solution.class │ ├── _059_Spiral_Matrix_II │ └── Solution.class │ ├── _062_Unique_Paths │ └── Solution.class │ ├── _064_Minimum_Path_Sum │ └── Solution.class │ ├── _067_Add_Binary │ └── Solution.class │ ├── _069_Sqrt_x │ └── Solution.class │ ├── _070_Climbing_Stairs │ └── Solution.class │ ├── _072_Edit_Distance │ └── Solution.class │ ├── _075_Sort_Colors │ └── Solution.class │ ├── _076_Minimum_Window_Substring │ └── Solution.class │ ├── _077_Combinations │ └── Solution.class │ ├── _078_Subsets │ └── Solution.class │ ├── _079_Word_Search │ └── Solution.class │ ├── _083_Remove_Duplicates_from_Sorted_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _084_Largest_Rectangle_in_Histogram │ └── Solution.class │ ├── _091_Decode_Ways │ └── Solution.class │ ├── _093_Restore_IP_Addresses │ └── Solution.class │ ├── _094_Binary_Tree_Inorder_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _095_Unique_Binary_Search_Trees_II │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _096_Unique_Binary_Search_Trees │ └── Solution.class │ ├── _098_Validate_Binary_Search_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _101_Symmetric_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _102_Binary_Tree_Level_Order_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _103_Binary_Tree_Zigzag_Level_Order_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _104_Maximum_Depth_of_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _108_Convert_Sorted_Array_to_Binary_Search_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _109_Convert_Sorted_List_to_Binary_Search_Tree │ ├── Solution$ListNode.class │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _110_Balanced_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _111_Minimum_Depth_of_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _112_Path_Sum │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _113_Path_Sum_II │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _114_Flatten_Binary_Tree_to_Linked_List │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _116_Populating_Next_Right_Pointers_in_Each_Node │ ├── Solution$Node.class │ └── Solution.class │ ├── _121_Best_Time_to_Buy_and_Sell_Stock │ └── Solution.class │ ├── _122_Best_Time_to_Buy_and_Sell_Stock_II │ └── Solution.class │ ├── _123_Best_Time_to_Buy_and_Sell_Stock_III │ └── Solution.class │ ├── _124_Binary_Tree_Maximum_Path_Sum │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _125_Valid_Palindrome │ └── Solution.class │ ├── _127_Word_Ladder │ └── Solution.class │ ├── _128_Longest_Consecutive_Sequence │ └── Solution.class │ ├── _130_Surrounded_Regions │ └── Solution.class │ ├── _131_Palindrome_Partitioning │ └── Solution.class │ ├── _134_Gas_Station │ └── Solution.class │ ├── _135_Candy │ └── Solution.class │ ├── _136_Single_Number │ └── Solution.class │ ├── _138_Copy_List_with_Random_Pointer │ ├── Solution$Node.class │ └── Solution.class │ ├── _139_Word_Break │ └── Solution.class │ ├── _140_Word_Break_II │ └── Solution.class │ ├── _141_Linked_List_Cycle │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _142_Linked_List_Cycle_II │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _143_Reorder_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _144_Binary_Tree_Preorder_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _145_Binary_Tree_Postorder_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _148_Sort_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _152_Maximum_Product_Subarray │ └── Solution.class │ ├── _153_Find_Minimum_in_Rotated_Sorted_Array │ └── Solution.class │ ├── _154_Find_Minimum_in_Rotated_Sorted_Array_II │ └── Solution.class │ ├── _155_Min_Stack │ └── MinStack.class │ ├── _160_Intersection_of_Two_Linked_Lists │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _169_Majority_Element │ └── Solution.class │ ├── _179_Largest_Number │ ├── Solution$1.class │ └── Solution.class │ ├── _190_Reverse_Bits │ └── Solution.class │ ├── _198_House_Robber │ └── Solution.class │ ├── _199_Binary_Tree_Right_Side_View │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _200_Number_of_Islands │ ├── Solution.class │ ├── Solution2.class │ └── UnionFind.class │ ├── _202_Happy_Number │ └── Solution.class │ ├── _205_Isomorphic_Strings │ └── Solution.class │ ├── _206_Reverse_Linked_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _207_Course_Schedule │ └── Solution.class │ ├── _208_Implement_Trie_Prefix_Tree │ ├── Solution$Trie.class │ ├── Solution$TrieNode.class │ └── Solution.class │ ├── _213_House_Robber2 │ └── Solution.class │ ├── _215_Kth_Largest_Element_in_an_Array │ └── Solution.class │ ├── _216_Combination_Sum_III │ └── Solution.class │ ├── _221_Maximal_Square │ └── Solution.class │ ├── _222_Count_Complete_Tree_Nodes │ └── Solution.class │ ├── _225_Implement_Stack_using_Queues │ └── MyStack.class │ ├── _226_Invert_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _230_Kth_Smallest_Element_in_a_BST │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _232_Implement_Queue_using_Stacks │ └── MyQueue.class │ ├── _234_Palindrome_Linked_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _235_Lowest_Common_Ancestor_of_a_Binary_Search_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _236_Lowest_Common_Ancestor_of_a_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _238_Product_of_Array_Except_Self │ └── Solution.class │ ├── _239_Sliding_Window_Maximum │ └── Solution.class │ ├── _240_Search_a_2D_Matrix_II │ └── Solution.class │ ├── _242_Valid_Anagram │ └── Solution.class │ ├── _257_Binary_Tree_Paths │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _260_Single_Number_III │ └── Solution.class │ ├── _263_Ugly_Number │ └── Solution.class │ ├── _264_Ugly_Number_II │ └── Solution.class │ ├── _268_Missing_Number │ └── Solution.class │ ├── _279_Perfect_Squares │ └── Solution.class │ ├── _283_Move_Zeroes │ └── Solution.class │ ├── _287_Find_the_Duplicate_Number │ └── Solution.class │ ├── _295_Find_Median_from_Data_Stream │ └── MedianFinder.class │ ├── _297_Serialize_and_Deserialize_Binary_Tree │ ├── Codec$TreeNode.class │ └── Codec.class │ ├── _300_Longest_Increasing_Subsequence │ └── Solution.class │ ├── _301_Remove_Invalid_Parentheses │ └── Solution.class │ ├── _303_Range_Sum_Query_Immutable │ └── NumArray.class │ ├── _309_Best_Time_to_Buy_and_Sell_Stock_with_Cooldown │ └── Solution.class │ ├── _312_Burst_Balloons │ └── Solution.class │ ├── _322_Coin_Change │ └── Solution.class │ ├── _328_Odd_Even_Linked_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _329_Longest_Increasing_Path_in_a_Matrix │ └── Solution.class │ ├── _334_Increasing_Triplet_Subsequence │ └── Solution.class │ ├── _337_House_Robber_III │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _338_Counting_Bits │ └── Solution.class │ ├── _343_Integer_Break │ └── Solution.class │ ├── _344_Reverse_String │ └── Solution.class │ ├── _347_Top_K_Frequent_Elements │ └── Solution.class │ ├── _350_Intersection_of_Two_Arrays_II │ └── Solution.class │ ├── _376_Wiggle_Subsequence │ └── Solution.class │ ├── _377_Combination_Sum_IV │ └── Solution.class │ ├── _378_Kth_Smallest_Element_in_a_Sorted_Matrix │ └── Solution.class │ ├── _387_First_Unique_Character_in_a_String │ └── Solution.class │ ├── _392_Is_Subsequence │ └── Solution.class │ ├── _394_Decode_String │ └── Solution.class │ ├── _395_Longest_Substring_with_At_Least_K_Repeating_Characters │ └── Solution.class │ ├── _404_Sum_of_Left_Leaves │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _405_Convert_a_Number_to_Hexadecimal │ └── Solution.class │ ├── _406_Queue_Reconstruction_by_Height │ ├── Solution$1.class │ └── Solution.class │ ├── _409_Longest_Palindrome │ └── Solution.class │ ├── _413_Arithmetic_Slices │ └── Solution.class │ ├── _416_Partition_Equal_Subset_Sum │ └── Solution.class │ ├── _435_Non_overlapping_Intervals │ ├── Solution$1.class │ ├── Solution$Interval.class │ └── Solution.class │ ├── _437_Path_Sum_III │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _438_Find_All_Anagrams_in_a_String │ └── Solution.class │ ├── _445_Add_Two_Numbers_II │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _448_Find_All_Numbers_Disappeared_in_an_Array │ └── Solution.class │ ├── _452_Minimum_Number_of_Arrows_to_Burst_Balloons │ ├── Solution$1.class │ └── Solution.class │ ├── _455_Assign_Cookies │ └── Solution.class │ ├── _460_LFU_Cache │ └── LFUCache.class │ ├── _461_Hamming_Distance │ └── Solution.class │ ├── _462_Minimum_Moves_to_Equal_Array_Elements_II │ └── Solution.class │ ├── _474_Ones_and_Zeroes │ └── Solution.class │ ├── _485_Max_Consecutive_Ones │ └── Solution.class │ ├── _494_Target_Sum │ └── Solution.class │ ├── _496_Next_Greater_Element_I │ └── Solution.class │ ├── _501_Find_Mode_in_Binary_Search_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _503_Next_Greater_Element_II │ └── Solution.class │ ├── _504_Base_7 │ └── Solution.class │ ├── _513_Find_Bottom_Left_Tree_Value │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _518_Coin_Change_2 │ └── Solution.class │ ├── _524_Longest_Word_in_Dictionary_through_Deleting │ └── Solution.class │ ├── _530_Minimum_Absolute_Difference_in_BST │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _538_Convert_BST_to_Greater_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _540_Single_Element_in_a_Sorted_Array │ └── Solution.class │ ├── _543_Diameter_of_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _547_Friend_Circles │ └── Solution.class │ ├── _557_Reverse_Words_in_a_String_III │ └── Solution.class │ ├── _560_Subarray_Sum_Equals_K │ └── Solution.class │ ├── _565_Array_Nesting │ └── Solution.class │ ├── _566_Reshape_the_Matrix │ └── Solution.class │ ├── _572_Subtree_of_Another_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _581_Shortest_Unsorted_Continuous_Subarray │ └── Solution.class │ ├── _583_Delete_Operation_for_Two_Strings │ └── Solution.class │ ├── _605_Can_Place_Flowers │ └── Solution.class │ ├── _617_Merge_Two_Binary_Trees │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _621_Task_Scheduler │ └── Solution.class │ ├── _637_Average_of_Levels_in_Binary_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _645_Set_Mismatch │ └── Solution.class │ ├── _646_Maximum_Length_of_Pair_Chain │ └── Solution.class │ ├── _647_Palindromic_Substrings │ └── Solution.class │ ├── _650_2_Keys_Keyboard │ └── Solution.class │ ├── _653_Two_Sum_IV_Input_is_a_BST │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _665_Non_decreasing_Array │ └── Solution.class │ ├── _669_Trim_a_Binary_Search_Tree │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _680_Valid_Palindrome_II │ └── Solution.class │ ├── _687_Longest_Univalue_Path │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _695_Max_Area_of_Island │ └── Solution.class │ ├── _697_Degree_of_an_Array │ └── Solution.class │ ├── _714_Best_Time_to_Buy_and_Sell_Stock_with_Transaction_Fee │ └── Solution.class │ ├── _725_Split_Linked_List_in_Parts │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _739_Daily_Temperatures │ └── Solution.class │ ├── _746_Min_Cost_Climbing_Stairs │ └── Solution.class │ ├── _763_Partition_Labels │ └── Solution.class │ ├── _769_Max_Chunks_To_Make_Sorted │ └── Solution.class │ ├── _771_Jewels_and_Stones │ └── Solution.class │ ├── _876_Middle_of_the_Linked_List │ ├── Solution$ListNode.class │ └── Solution.class │ ├── _889_Construct_Binary_Tree_from_Preorder_and_Postorder_Traversal │ ├── Solution$TreeNode.class │ └── Solution.class │ ├── _90_Subsets_II │ └── Solution.class │ └── _946_Validate_Stack_Sequences │ └── Solution.class └── src └── algorithm ├── _001_Two_Sum └── Solution.java ├── _002_Add_Two_Numbers └── Solution.java ├── _003_Longest_Substring_Without_Repeating_Characters └── Solution.java ├── _004_Median_of_Two_Sorted_Arrays └── Solution.java ├── _005_Longest_Palindromic_Substring └── Solution.java ├── _007_Reverse_Integer └── Solution.java ├── _008_String_to_Integer └── Solution.java ├── _009_Palindrome_Number └── Solution.java ├── _011_Container_With_Most_Water └── Solution.java ├── _012_Integer_to_Roman └── Solution.java ├── _013_Roman_to_Integer └── Solution.java ├── _014_Longest_Common_Prefix └── Solution.java ├── _015_3Sum └── Solution.java ├── _016_3Sum_Closest └── Solution.java ├── _017_Letter_Combinations_of_a_Phone_Number └── Solution.java ├── _019_Remove_Nth_Node_From_End_of_List └── Solution.java ├── _020_Valid_Parentheses └── Solution.java ├── _021_Merge_Two_Sorted_Lists └── Solution.java ├── _022_Generate_Parentheses └── Solution.java ├── _023_Merge_k_Sorted_Lists └── Solution.java ├── _024_Swap_Nodes_in_Pairs └── Solution.java ├── _025_Reverse_Nodes_in_k_Group └── Solution.java ├── _026_Remove_Duplicates_from_Sorted_Array └── Solution.java ├── _027_Remove_Element └── Solution.java ├── _028_Implement_strStr └── Solution.java ├── _031_Next_Permutation └── Solution.java ├── _032_Longest_Valid_Parentheses └── Solution.java ├── _033_Search_in_Rotated_Sorted_Array └── Solution.java ├── _034_Find_First_and_Last_Position_of_Element_in_Sorted_Array └── Solution.java ├── _035_Search_Insert_Position └── Solution.java ├── _038_Count_and_Say └── Solution.java ├── _039_Combination_Sum └── Solution.java ├── _040_Combination_Sum_ll └── Solution.java ├── _041_First_Missing_Positive └── Solution.java ├── _042_Trapping_Rain_Water └── Solution.java ├── _043_Multiply_Strings └── Solution.java ├── _045_Jump_Game_II └── Solution.java ├── _046_Permutations ├── Solution.java └── Solution2.java ├── _047_Permutations_II ├── Solution.java └── Solution2.java ├── _048_Rotate_Image └── Solution.java ├── _049_Group_Anagrams └── Solution.java ├── _050_Pow └── Solution.java ├── _053_Maximum_Subarray └── Solution.java ├── _054_Spiral_Matrix └── Solution.java ├── _055_Jump_Game └── Solution.java ├── _056_Merge_Intervals └── Solution.java ├── _059_Spiral_Matrix_II └── Solution.java ├── _062_Unique_Paths └── Solution.java ├── _064_Minimum_Path_Sum └── Solution.java ├── _067_Add_Binary └── Solution.java ├── _069_Sqrt_x └── Solution.java ├── _070_Climbing_Stairs └── Solution.java ├── _072_Edit_Distance └── Solution.java ├── _075_Sort_Colors └── Solution.java ├── _076_Minimum_Window_Substring └── Solution.java ├── _077_Combinations └── Solution.java ├── _078_Subsets └── Solution.java ├── _079_Word_Search └── Solution.java ├── _083_Remove_Duplicates_from_Sorted_List └── Solution.java ├── _084_Largest_Rectangle_in_Histogram └── Solution.java ├── _088_Merge_Sorted_Array └── Solution.java ├── _090_Subsets_II └── Solution.java ├── _091_Decode_Ways └── Solution.java ├── _093_Restore_IP_Addresses └── Solution.java ├── _094_Binary_Tree_Inorder_Traversal └── Solution.java ├── _095_Unique_Binary_Search_Trees_II └── Solution.java ├── _096_Unique_Binary_Search_Trees └── Solution.java ├── _098_Validate_Binary_Search_Tree └── Solution.java ├── _101_Symmetric_Tree └── Solution.java ├── _102_Binary_Tree_Level_Order_Traversal └── Solution.java ├── _103_Binary_Tree_Zigzag_Level_Order_Traversal └── Solution.java ├── _104_Maximum_Depth_of_Binary_Tree └── Solution.java ├── _105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal └── Solution.java ├── _108_Convert_Sorted_Array_to_Binary_Search_Tree └── Solution.java ├── _109_Convert_Sorted_List_to_Binary_Search_Tree └── Solution.java ├── _110_Balanced_Binary_Tree └── Solution.java ├── _111_Minimum_Depth_of_Binary_Tree └── Solution.java ├── _112_Path_Sum └── Solution.java ├── _113_Path_Sum_II └── Solution.java ├── _114_Flatten_Binary_Tree_to_Linked_List └── Solution.java ├── _116_Populating_Next_Right_Pointers_in_Each_Node └── Solution.java ├── _121_Best_Time_to_Buy_and_Sell_Stock └── Solution.java ├── _122_Best_Time_to_Buy_and_Sell_Stock_II └── Solution.java ├── _123_Best_Time_to_Buy_and_Sell_Stock_III └── Solution.java ├── _124_Binary_Tree_Maximum_Path_Sum └── Solution.java ├── _125_Valid_Palindrome └── Solution.java ├── _127_Word_Ladder └── Solution.java ├── _128_Longest_Consecutive_Sequence └── Solution.java ├── _130_Surrounded_Regions └── Solution.java ├── _131_Palindrome_Partitioning └── Solution.java ├── _134_Gas_Station └── Solution.java ├── _135_Candy └── Solution.java ├── _136_Single_Number └── Solution.java ├── _138_Copy_List_with_Random_Pointer └── Solution.java ├── _139_Word_Break └── Solution.java ├── _140_Word_Break_II └── Solution.java ├── _141_Linked_List_Cycle └── Solution.java ├── _142_Linked_List_Cycle_II └── Solution.java ├── _143_Reorder_List └── Solution.java ├── _144_Binary_Tree_Preorder_Traversal └── Solution.java ├── _145_Binary_Tree_Postorder_Traversal └── Solution.java ├── _148_Sort_List └── Solution.java ├── _152_Maximum_Product_Subarray └── Solution.java ├── _153_Find_Minimum_in_Rotated_Sorted_Array └── Solution.java ├── _154_Find_Minimum_in_Rotated_Sorted_Array_II └── Solution.java ├── _155_Min_Stack └── MinStack.java ├── _160_Intersection_of_Two_Linked_Lists └── Solution.java ├── _169_Majority_Element └── Solution.java ├── _179_Largest_Number └── Solution.java ├── _190_Reverse_Bits └── Solution.java ├── _198_House_Robber └── Solution.java ├── _199_Binary_Tree_Right_Side_View └── Solution.java ├── _200_Number_of_Islands ├── Solution.java └── Solution2.java ├── _202_Happy_Number └── Solution.java ├── _205_Isomorphic_Strings └── Solution.java ├── _206_Reverse_Linked_List └── Solution.java ├── _207_Course_Schedule └── Solution.java ├── _208_Implement_Trie_Prefix_Tree └── Solution.java ├── _213_House_Robber2 └── Solution.java ├── _215_Kth_Largest_Element_in_an_Array └── Solution.java ├── _216_Combination_Sum_III └── Solution.java ├── _221_Maximal_Square └── Solution.java ├── _222_Count_Complete_Tree_Nodes └── Solution.java ├── _225_Implement_Stack_using_Queues └── MyStack.java ├── _226_Invert_Binary_Tree └── Solution.java ├── _230_Kth_Smallest_Element_in_a_BST └── Solution.java ├── _232_Implement_Queue_using_Stacks └── MyQueue.java ├── _234_Palindrome_Linked_List └── Solution.java ├── _235_Lowest_Common_Ancestor_of_a_Binary_Search_Tree └── Solution.java ├── _236_Lowest_Common_Ancestor_of_a_Binary_Tree └── Solution.java ├── _238_Product_of_Array_Except_Self └── Solution.java ├── _239_Sliding_Window_Maximum └── Solution.java ├── _240_Search_a_2D_Matrix_II └── Solution.java ├── _242_Valid_Anagram └── Solution.java ├── _257_Binary_Tree_Paths └── Solution.java ├── _260_Single_Number_III └── Solution.java ├── _263_Ugly_Number └── Solution.java ├── _264_Ugly_Number_II └── Solution.java ├── _268_Missing_Number └── Solution.java ├── _279_Perfect_Squares └── Solution.java ├── _283_Move_Zeroes └── Solution.java ├── _287_Find_the_Duplicate_Number └── Solution.java ├── _295_Find_Median_from_Data_Stream └── MedianFinder.java ├── _297_Serialize_and_Deserialize_Binary_Tree └── Codec.java ├── _300_Longest_Increasing_Subsequence └── Solution.java ├── _301_Remove_Invalid_Parentheses └── Solution.java ├── _303_Range_Sum_Query_Immutable └── NumArray.java ├── _309_Best_Time_to_Buy_and_Sell_Stock_with_Cooldown └── Solution.java ├── _312_Burst_Balloons └── Solution.java ├── _322_Coin_Change └── Solution.java ├── _328_Odd_Even_Linked_List └── Solution.java ├── _329_Longest_Increasing_Path_in_a_Matrix └── Solution.java ├── _334_Increasing_Triplet_Subsequence └── Solution.java ├── _337_House_Robber_III └── Solution.java ├── _338_Counting_Bits └── Solution.java ├── _343_Integer_Break └── Solution.java ├── _344_Reverse_String └── Solution.java ├── _347_Top_K_Frequent_Elements └── Solution.java ├── _350_Intersection_of_Two_Arrays_II └── Solution.java ├── _376_Wiggle_Subsequence └── Solution.java ├── _377_Combination_Sum_IV └── Solution.java ├── _378_Kth_Smallest_Element_in_a_Sorted_Matrix └── Solution.java ├── _387_First_Unique_Character_in_a_String └── Solution.java ├── _392_Is_Subsequence └── Solution.java ├── _394_Decode_String └── Solution.java ├── _395_Longest_Substring_with_At_Least_K_Repeating_Characters └── Solution.java ├── _404_Sum_of_Left_Leaves └── Solution.java ├── _405_Convert_a_Number_to_Hexadecimal └── Solution.java ├── _406_Queue_Reconstruction_by_Height └── Solution.java ├── _409_Longest_Palindrome └── Solution.java ├── _413_Arithmetic_Slices └── Solution.java ├── _416_Partition_Equal_Subset_Sum └── Solution.java ├── _435_Non_overlapping_Intervals └── Solution.java ├── _437_Path_Sum_III └── Solution.java ├── _438_Find_All_Anagrams_in_a_String └── Solution.java ├── _445_Add_Two_Numbers_II └── Solution.java ├── _448_Find_All_Numbers_Disappeared_in_an_Array └── Solution.java ├── _452_Minimum_Number_of_Arrows_to_Burst_Balloons └── Solution.java ├── _455_Assign_Cookies └── Solution.java ├── _460_LFU_Cache └── LFUCache.java ├── _461_Hamming_Distance └── Solution.java ├── _462_Minimum_Moves_to_Equal_Array_Elements_II └── Solution.java ├── _474_Ones_and_Zeroes └── Solution.java ├── _485_Max_Consecutive_Ones └── Solution.java ├── _494_Target_Sum └── Solution.java ├── _496_Next_Greater_Element_I └── Solution.java ├── _501_Find_Mode_in_Binary_Search_Tree └── Solution.java ├── _503_Next_Greater_Element_II └── Solution.java ├── _504_Base_7 └── Solution.java ├── _513_Find_Bottom_Left_Tree_Value └── Solution.java ├── _518_Coin_Change_2 └── Solution.java ├── _524_Longest_Word_in_Dictionary_through_Deleting └── Solution.java ├── _530_Minimum_Absolute_Difference_in_BST └── Solution.java ├── _538_Convert_BST_to_Greater_Tree └── Solution.java ├── _540_Single_Element_in_a_Sorted_Array └── Solution.java ├── _543_Diameter_of_Binary_Tree └── Solution.java ├── _547_Friend_Circles └── Solution.java ├── _557_Reverse_Words_in_a_String_III └── Solution.java ├── _560_Subarray_Sum_Equals_K └── Solution.java ├── _565_Array_Nesting └── Solution.java ├── _566_Reshape_the_Matrix └── Solution.java ├── _572_Subtree_of_Another_Tree └── Solution.java ├── _581_Shortest_Unsorted_Continuous_Subarray └── Solution.java ├── _583_Delete_Operation_for_Two_Strings └── Solution.java ├── _605_Can_Place_Flowers └── Solution.java ├── _617_Merge_Two_Binary_Trees └── Solution.java ├── _621_Task_Scheduler └── Solution.java ├── _637_Average_of_Levels_in_Binary_Tree └── Solution.java ├── _645_Set_Mismatch └── Solution.java ├── _646_Maximum_Length_of_Pair_Chain └── Solution.java ├── _647_Palindromic_Substrings └── Solution.java ├── _650_2_Keys_Keyboard └── Solution.java ├── _653_Two_Sum_IV_Input_is_a_BST └── Solution.java ├── _665_Non_decreasing_Array └── Solution.java ├── _669_Trim_a_Binary_Search_Tree └── Solution.java ├── _680_Valid_Palindrome_II └── Solution.java ├── _687_Longest_Univalue_Path └── Solution.java ├── _695_Max_Area_of_Island └── Solution.java ├── _697_Degree_of_an_Array └── Solution.java ├── _714_Best_Time_to_Buy_and_Sell_Stock_with_Transaction_Fee └── Solution.java ├── _725_Split_Linked_List_in_Parts └── Solution.java ├── _739_Daily_Temperatures └── Solution.java ├── _746_Min_Cost_Climbing_Stairs └── Solution.java ├── _763_Partition_Labels └── Solution.java ├── _769_Max_Chunks_To_Make_Sorted └── Solution.java ├── _771_Jewels_and_Stones └── Solution.java ├── _876_Middle_of_the_Linked_List └── Solution.java ├── _889_Construct_Binary_Tree_from_Preorder_and_Postorder_Traversal └── Solution.java └── _946_Validate_Stack_Sequences └── Solution.java /.idea/misc.xml: -------------------------------------------------------------------------------- 1 | 2 | 3 | 4 | 5 | 6 | -------------------------------------------------------------------------------- /.idea/modules.xml: 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nums[i]); 23 | return result; 24 | } 25 | map.put(nums[i], i); 26 | } 27 | return result; 28 | } 29 | 30 | 31 | } 32 | -------------------------------------------------------------------------------- /src/algorithm/_007_Reverse_Integer/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._007_Reverse_Integer; 2 | 3 | public class Solution { 4 | 5 | /*反转整数*/ 6 | 7 | 8 | public int reverse(int x) { 9 | int result = 0; 10 | while (x != 0) { 11 | int tail = x % 10; 12 | int newResult = result * 10 + tail; 13 | //判断是否溢出 14 | if ((newResult - tail) / 10 != result) { 15 | return 0; 16 | } 17 | result = newResult; 18 | x /= 10; 19 | } 20 | return result; 21 | } 22 | } 23 | -------------------------------------------------------------------------------- /src/algorithm/_009_Palindrome_Number/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._009_Palindrome_Number; 2 | 3 | public class Solution { 4 | /*判断是否是回文数*/ 5 | public boolean isPalindromeOther(int x) { 6 | if (x < 0 || x % 10 == 0 && x != 0) { 7 | return false; 8 | } 9 | int reverse = 0; 10 | while (x > reverse) { 11 | reverse = reverse * 10 + x % 10; 12 | x = x / 10; 13 | } 14 | return x == reverse || x == reverse / 10; 15 | } 16 | } 17 | -------------------------------------------------------------------------------- /src/algorithm/_011_Container_With_Most_Water/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._011_Container_With_Most_Water; 2 | 3 | public class Solution { 4 | 5 | /*给定 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线， 6 | 垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。*/ 7 | //示例图片在images文件夹 8 | 9 | //定义双指针 10 | public int maxArea(int[] height) { 11 | int left = 0; 12 | int right = height.length - 1; 13 | int max = 0; 14 | while (left < right) { 15 | //求最大面积 16 | max = Math.max(max, Math.min(height[left], height[right]) * (right - left)); 17 | //如果右边大，移动左指针。如果左边大，移动右指针 18 | if (height[left] < height[right]) { 19 | left++; 20 | } else { 21 | right--; 22 | } 23 | } 24 | return max; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_019_Remove_Nth_Node_From_End_of_List/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._019_Remove_Nth_Node_From_End_of_List; 2 | 3 | public class Solution { 4 | 5 | public class ListNode { 6 | int val; 7 | ListNode next; 8 | ListNode(int x) { val = x; } 9 | } 10 | 11 | 12 | /*删除链表倒数第n个节点*/ 13 | 14 | public ListNode removeNthFromEnd(ListNode head, int n) { 15 | //构造一个节点，并让它连接首节点，很关键 16 | ListNode start = new ListNode(0); 17 | start.next = head; 18 | ListNode fast = start,slow = start; 19 | //快指针先走n+1步 20 | for(int i = 0;i <= n;i++){ 21 | fast = fast.next; 22 | } 23 | //快慢指针一起走 24 | while(fast != null){ 25 | fast = fast.next; 26 | slow = slow.next; 27 | } 28 | //移除倒数第n个节点 29 | slow.next = slow.next.next; 30 | return start.next; 31 | } 32 | } 33 | -------------------------------------------------------------------------------- /src/algorithm/_021_Merge_Two_Sorted_Lists/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._021_Merge_Two_Sorted_Lists; 2 | 3 | public class Solution { 4 | 5 | //链表结点 6 | public class ListNode { 7 | int val; 8 | ListNode next; 9 | 10 | ListNode(int x) { 11 | val = x; 12 | } 13 | } 14 | 15 | //合并两个已排序的链接列表并将其作为新列表返回,新列表应该通过拼接前两个列表的节点来完成。 16 | public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 17 | //递归的结束条件 18 | if (l1 == null) return l2; 19 | if (l2 == null) return l1; 20 | if (l1.val < l2.val) { 21 | l1.next = mergeTwoLists(l1.next, l2); 22 | return l1; 23 | } else { 24 | l2.next = mergeTwoLists(l1, l2.next); 25 | return l2; 26 | } 27 | } 28 | } 29 | 30 | 31 | -------------------------------------------------------------------------------- /src/algorithm/_024_Swap_Nodes_in_Pairs/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._024_Swap_Nodes_in_Pairs; 2 | 3 | public class Solution { 4 | 5 | /*给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。 6 | 你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。*/ 7 | 8 | public class ListNode { 9 | int val; 10 | ListNode next; 11 | 12 | ListNode(int x) { 13 | val = x; 14 | } 15 | } 16 | 17 | 18 | public ListNode swapPairs(ListNode head) { 19 | if (head == null || head.next == null) { 20 | return head; 21 | } 22 | ListNode second = head.next; 23 | ListNode third = second.next; 24 | //改变结点的指向 25 | second.next = head; 26 | head.next = swapPairs(third); 27 | return second; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_026_Remove_Duplicates_from_Sorted_Array/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._026_Remove_Duplicates_from_Sorted_Array; 2 | 3 | public class Solution { 4 | 5 | /*给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素只出现一次，返回移除后数组的新长度。 6 | 不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。*/ 7 | 8 | public int removeDuplicates(int[] nums) { 9 | int count = 0; 10 | for (int i = 1; i < nums.length; i++) { 11 | //出现重复的数字，count次数加1 12 | if (nums[i] == nums[i - 1]) { 13 | count++; 14 | } else { 15 | nums[i - count] = nums[i]; 16 | } 17 | } 18 | //总次数减去出现重复的次数 19 | return nums.length - count; 20 | } 21 | } 22 | -------------------------------------------------------------------------------- /src/algorithm/_028_Implement_strStr/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._028_Implement_strStr; 2 | 3 | public class Solution { 4 | 5 | //判断字符串s是否包含了字符串t 6 | public int strStrOther(String s, String t) { 7 | if (t.isEmpty()) return 0; 8 | for (int i = 0; i <= s.length() - t.length(); i++) { 9 | //两个字符串开始相等的部分 10 | for (int j = 0; j < t.length() && s.charAt(i + j) == t.charAt(j); j++) 11 | //如果j到了字符串的最后一位，就可以说明包含了 12 | if (j == t.length() - 1) return i; 13 | } 14 | return -1; 15 | } 16 | 17 | 18 | } 19 | -------------------------------------------------------------------------------- /src/algorithm/_032_Longest_Valid_Parentheses/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._032_Longest_Valid_Parentheses; 2 | 3 | import java.util.Stack; 4 | 5 | public class Solution { 6 | 7 | /*给定一个只包含 '(' 和 ')' 的字符串，找出最长的包含有效括号的子串的长度。*/ 8 | 9 | //借助栈保存s中字符为‘（’的位置 10 | public int longestValidParentheses(String s) { 11 | Stack stack = new Stack<>(); 12 | int result = 0; 13 | //-1用于后续计算长度 14 | stack.push(-1); 15 | for (int i = 0; i < s.length(); i++) { 16 | //如果当前字符为）且栈顶数字表示的字符为（.要限制栈的大小大于1，不然一开始插入的-1会导致s.charAt()报错 17 | if (s.charAt(i) == ')' && stack.size() > 1 && s.charAt(stack.peek()) == '(') { 18 | //将(出栈 19 | stack.pop(); 20 | //更新有效括号子串的最长长度 21 | result = Math.max(result, i - stack.peek()); 22 | } else { 23 | stack.push(i); 24 | } 25 | } 26 | return result; 27 | } 28 | 29 | //todo 动态规划 30 | } 31 | -------------------------------------------------------------------------------- /src/algorithm/_035_Search_Insert_Position/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._035_Search_Insert_Position; 2 | 3 | public class Solution { 4 | 5 | /*给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。 6 | 如果目标值不存在于数组中，返回它将会被按顺序插入的位置。 7 | 你可以假设数组中无重复元素。*/ 8 | 9 | 10 | //二分查找 11 | public int searchInsert(int[] nums, int target) { 12 | int start = 0, end = nums.length - 1; 13 | while (start <= end) { 14 | int mid = start + (end - start) / 2; 15 | if (nums[mid] == target) { 16 | return mid; 17 | } else if (nums[mid] > target) { 18 | end = mid - 1; 19 | } else { 20 | start = mid + 1; 21 | } 22 | } 23 | return start; 24 | } 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_041_First_Missing_Positive/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._041_First_Missing_Positive; 2 | 3 | public class Solution { 4 | 5 | /*给定一个未排序的整数数组，找出其中没有出现的最小的正整数。*/ 6 | 7 | //时间复杂度要求o(n) 8 | //算法的思想就是让数组里面存放着1-n的正数，第0个位置放1，第1个位置放2，以此类推 9 | // 最后遍历数组如果没有存放匹配的正数，返回最小正数的值为该位置+1 10 | public int firstMissingPositive(int[] nums) { 11 | for (int i = 0; i < nums.length; i++) { 12 | //如果满足下述匹配规则，就一直交换 13 | while (nums[i] > 0 && nums[i] <= nums.length && nums[nums[i] - 1] != nums[i]) { 14 | swap(nums, i, nums[i] - 1); 15 | } 16 | 17 | } 18 | //遍历数组找出不匹配的 19 | for (int i = 0; i < nums.length; i++) { 20 | if (nums[i] != i + 1) { 21 | return i + 1; 22 | } 23 | } 24 | //表示数组完全匹配规则，最小正数为数组长度+1 25 | return nums.length + 1; 26 | } 27 | 28 | private void swap(int[] nums, int i, int j) { 29 | int temp = nums[i]; 30 | nums[i] = nums[j]; 31 | nums[j] = temp; 32 | } 33 | } 34 | -------------------------------------------------------------------------------- /src/algorithm/_045_Jump_Game_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._045_Jump_Game_II; 2 | 3 | public class Solution { 4 | 5 | /*给定一个非负整数数组，你最初位于数组的第一个位置。 6 | 数组中的每个元素代表你在该位置可以跳跃的最大长度。 7 | 你的目标是使用最少的跳跃次数到达数组的最后一个位置。 8 | */ 9 | 10 | public int jump(int[] nums) { 11 | int jump = 0, curEnd = 0, curFarthest = 0; 12 | for (int i = 0; i < nums.length - 1; i++) { 13 | //每次更新跳到最远的位置 14 | curFarthest = Math.max(curFarthest, nums[i] + i); 15 | //如果等于当前结束值，步数增加1，并且更新当前结束值等于过程中更新的最远的位置 16 | if (i == curEnd) { 17 | jump++; 18 | curEnd = Math.max(curFarthest, curEnd); 19 | } 20 | } 21 | return jump; 22 | } 23 | } 24 | -------------------------------------------------------------------------------- /src/algorithm/_048_Rotate_Image/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._048_Rotate_Image; 2 | 3 | public class Solution { 4 | 5 | //将图像顺时针旋转90度 6 | 7 | 8 | //图形变化如下 9 | // 1 2 3 1 4 7 7 4 1 10 | // 4 5 6 2 5 8 8 5 2 11 | // 7 8 9 3 6 9 9 6 3 12 | 13 | public void rotate(int[][] matrix) { 14 | for (int i = 0; i < matrix.length; i++) { 15 | for (int j = i + 1; j < matrix.length; j++) { 16 | int temp = matrix[j][i]; 17 | matrix[j][i] = matrix[i][j]; 18 | matrix[i][j] = temp; 19 | } 20 | } 21 | for (int i = 0; i < matrix.length; i++) { 22 | for (int j = 0; j < matrix.length / 2; j++) { 23 | int temp = matrix[i][j]; 24 | matrix[i][j] = matrix[i][matrix.length - 1 - j]; 25 | matrix[i][matrix.length - 1 - j] = temp; 26 | } 27 | } 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_050_Pow/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._050_Pow; 2 | 3 | public class Solution { 4 | //我一开始用for循环，time limit，可能计算量太大,而且方法太暴力。 5 | //使用递归才是正确做法， 6 | public double myPow(double x, int n) { 7 | if (n == 0) 8 | return 1; 9 | if (n < 0) { 10 | //如果n是最小值的话就要把它变大，不然加负号会溢出 11 | if (n == Integer.MIN_VALUE) { 12 | n /= 2; 13 | x *= x; 14 | } 15 | n = -n; 16 | x = 1 / x; 17 | } 18 | //递归调用，底数平方，指数减半 19 | return (n % 2 == 0) ? myPow(x * x, n / 2) : x * myPow(x * x, n / 2); 20 | } 21 | } 22 | -------------------------------------------------------------------------------- /src/algorithm/_055_Jump_Game/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._055_Jump_Game; 2 | 3 | public class Solution { 4 | 5 | /*给定一个数组，初始定位在第一个索引，数组中的每个元素值代表你能跳到的最大长度， 6 | 判断是否能跳到最后一个位置*/ 7 | 8 | public boolean canJump(int[] nums) { 9 | //能跳到的最大位置 10 | int max = 0; 11 | for (int i = 0; i < nums.length; i++) { 12 | //如果最大位置比i小，表示跳不到 13 | if (i > max) { 14 | return false; 15 | } 16 | //每一次都更新能跳到的最大位置 17 | max = Math.max(max, nums[i] + i); 18 | } 19 | return true; 20 | } 21 | 22 | } 23 | -------------------------------------------------------------------------------- /src/algorithm/_062_Unique_Paths/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._062_Unique_Paths; 2 | 3 | public class Solution { 4 | 5 | //统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。 6 | 7 | public int uniquePaths(int m, int n) { 8 | int[][] dp = new int[m][n]; 9 | for (int i = 0; i < m; i++) { 10 | dp[i][0] = 1; 11 | } 12 | for (int j = 0; j < n; j++) { 13 | dp[0][j] = 1; 14 | } 15 | for (int i = 1; i < m; i++) { 16 | for (int j = 1; j < n; j++) { 17 | dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; 18 | } 19 | } 20 | return dp[m - 1][n - 1]; 21 | } 22 | } 23 | -------------------------------------------------------------------------------- /src/algorithm/_064_Minimum_Path_Sum/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._064_Minimum_Path_Sum; 2 | 3 | public class Solution { 4 | 5 | //求从矩阵的左上角到右下角的最小路径和，每次只能向右和向下移动。 6 | public int minPathSum(int[][] grid) { 7 | int m = grid.length, n = grid[0].length; 8 | for (int i = 0; i < m; i++) { 9 | for (int j = 0; j < n; j++) { 10 | if (i == 0 && j != 0) { 11 | grid[i][j] += grid[i][j - 1]; 12 | } 13 | if (i != 0 && j == 0) { 14 | grid[i][j] += grid[i - 1][j]; 15 | } 16 | if (i != 0 && j != 0) { 17 | grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]); 18 | } 19 | } 20 | } 21 | return grid[m - 1][n - 1]; 22 | } 23 | } 24 | -------------------------------------------------------------------------------- /src/algorithm/_067_Add_Binary/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._067_Add_Binary; 2 | 3 | public class Solution { 4 | 5 | //题目描述：给定两个二进制字符串，相加后返回字符串 6 | 7 | public String addBinary(String a, String b) { 8 | StringBuilder sb = new StringBuilder(); 9 | int i = a.length() - 1, j = b.length() - 1, carry = 0, sum; 10 | //如果字符串长度大于0就进入循环 11 | while (i >= 0 || j >= 0) { 12 | sum = carry; 13 | if (i >= 0) sum += a.charAt(i--) - '0'; 14 | if (j >= 0) sum += b.charAt(j--) - '0'; 15 | //添加二进制取值 16 | sb.append(sum % 2); 17 | //有无进位 18 | carry = sum / 2; 19 | } 20 | //有进位才添加 21 | if (carry != 0) sb.append(carry); 22 | //我认为最关键是要能想到最后字符串反转 23 | return sb.reverse().toString(); 24 | } 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_069_Sqrt_x/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._069_Sqrt_x; 2 | 3 | public class Solution { 4 | 5 | /*实现 int sqrt(int x) 函数。 6 | 计算并返回 x 的平方根，其中 x 是非负整数。 7 | 由于返回类型是整数，结果只保留整数的部分，小数部分将被舍去。*/ 8 | 9 | 10 | //平方根一定处于1-x之间 11 | public int mySqrt(int x) { 12 | if (x <= 1) { 13 | return x; 14 | } 15 | int l = 1, h = x; 16 | //二分查找 17 | while (l <= h) { 18 | int mid = l + (h - l) / 2; 19 | int sqrt = x / mid; 20 | if (sqrt > mid) { 21 | l = mid + 1; 22 | } else if (sqrt < mid) { 23 | h = mid - 1; 24 | } else { 25 | return mid; 26 | } 27 | } 28 | //h比l要小，比如8是返回2，返回较小的。 29 | return h; 30 | } 31 | } 32 | -------------------------------------------------------------------------------- /src/algorithm/_070_Climbing_Stairs/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._070_Climbing_Stairs; 2 | 3 | public class Solution { 4 | 5 | 6 | /*假设你正在爬楼梯。需要 n 阶你才能到达楼顶。 7 | 每次你可以爬 1 或 2 个台阶。你有多少种不同的方法可以爬到楼顶呢？ 8 | 注意：给定 n 是一个正整数。*/ 9 | 10 | 11 | //斐波拉契的思路 12 | //想象成一个数组，第n个位置的方法等于第n-1个位置的方法加上第n-2个位置的方法 13 | //第n-1个位置加1步就到达第n个位置，第n-2个位置加两步就到达第n个位置，解法加起来不会重复 14 | public int climbStairsOther(int n) { 15 | // base cases 16 | if(n <= 0) return 0; 17 | if(n == 1) return 1; 18 | if(n == 2) return 2; 19 | //一步之前 20 | int one_step_before = 2; 21 | //两步之前 22 | int two_steps_before = 1; 23 | int all_ways = 0; 24 | 25 | for(int i=2; i= 0 && j >= 0) { 16 | if (nums1[i] > nums2[j]) { 17 | nums1[k--] = nums1[i--]; 18 | } else { 19 | nums1[k--] = nums2[j--]; 20 | } 21 | } 22 | while (j >= 0) { 23 | nums1[k--] = nums2[j--]; 24 | } 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_090_Subsets_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._090_Subsets_II; 2 | 3 | import java.util.ArrayList; 4 | import java.util.Arrays; 5 | import java.util.List; 6 | 7 | public class Solution { 8 | 9 | /*给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。 10 | 说明：解集不能包含重复的子集。*/ 11 | 12 | public List> subsetsWithDup(int[] nums) { 13 | List> result = new ArrayList<>(); 14 | Arrays.sort(nums); 15 | backtrack(result, new ArrayList(), nums, 0); 16 | return result; 17 | } 18 | 19 | private void backtrack(List> result, List list, int[] nums, int start) { 20 | if (!result.contains(list)) { 21 | result.add(new ArrayList<>(list)); 22 | } 23 | for (int i = start; i < nums.length; i++) { 24 | list.add(nums[i]); 25 | backtrack(result, list, nums, i + 1); 26 | list.remove(list.size() - 1); 27 | } 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_096_Unique_Binary_Search_Trees/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._096_Unique_Binary_Search_Trees; 2 | 3 | public class Solution { 4 | /*给定一个整数 n，求以 1 ... n 为节点组成的二叉搜索树有多少种？*/ 5 | 6 | /* 7 | G(n) = F(1, n) + F(2, n) + ... + F(n, n). 8 | G(0)=1, G(1)=1. 9 | F(i, n) = G(i-1) * G(n-i) 1 <= i <= n 10 | G(n) = G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0) 11 | 函数F(i,n)表示n个结点以i为根节点的二叉搜索树，函数G(n)表示n个结点组成二叉搜索树的种数 12 | */ 13 | public int numTrees(int n) { 14 | int[] G = new int[n + 1]; 15 | G[0] = 1; 16 | G[1] = 1; 17 | for (int i = 2; i < n + 1; i++) { 18 | //该循环表示G(i) = F(1, i) + F(2, i) + ... + F(i, i) 19 | for (int j = 1; j <= i; j++) { 20 | G[i] += G[j - 1] * G[i - j]; 21 | } 22 | } 23 | return G[n]; 24 | } 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_101_Symmetric_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._101_Symmetric_Tree; 2 | 3 | public class Solution { 4 | 5 | public class TreeNode { 6 | int val; 7 | TreeNode left; 8 | TreeNode right; 9 | TreeNode(int x) { val = x; } 10 | } 11 | 12 | 13 | /*给定一个二叉树，检查它是否是镜像对称的。*/ 14 | 15 | public boolean isSymmetric(TreeNode root) { 16 | if(root == null){ 17 | return true; 18 | } 19 | return isSymmetric(root.left,root.right); 20 | } 21 | 22 | 23 | private boolean isSymmetric(TreeNode t1,TreeNode t2){ 24 | if(t1 == null && t2 == null){ 25 | return true; 26 | } 27 | if(t1 == null || t2 == null){ 28 | return false; 29 | } 30 | if(t1.val != t2.val){ 31 | return false; 32 | } 33 | return isSymmetric(t1.right,t2.left)&&isSymmetric(t1.left,t2.right); 34 | } 35 | } 36 | -------------------------------------------------------------------------------- /src/algorithm/_104_Maximum_Depth_of_Binary_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._104_Maximum_Depth_of_Binary_Tree; 2 | 3 | public class Solution { 4 | 5 | /*给定一个二叉树，找出其最大深度。 6 | 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。*/ 7 | 8 | public class TreeNode { 9 | int val; 10 | TreeNode left; 11 | TreeNode right; 12 | 13 | TreeNode(int x) { 14 | val = x; 15 | } 16 | } 17 | 18 | 19 | public int maxDepth(TreeNode root) { 20 | if (root == null) { 21 | return 0; 22 | } 23 | return 1 + Math.max(maxDepth(root.left), maxDepth(root.right)); 24 | } 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_111_Minimum_Depth_of_Binary_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._111_Minimum_Depth_of_Binary_Tree; 2 | 3 | public class Solution { 4 | 5 | 6 | public class TreeNode { 7 | int val; 8 | TreeNode left; 9 | TreeNode right; 10 | 11 | TreeNode(int x) { 12 | val = x; 13 | } 14 | } 15 | 16 | /*给定一个二叉树，找到最小深度*/ 17 | 18 | public int minDepth(TreeNode root) { 19 | if (root == null) { 20 | return 0; 21 | } 22 | int left = minDepth(root.left); 23 | int right = minDepth(root.right); 24 | if (left == 0 || right == 0) { 25 | return left + right + 1; 26 | } 27 | return Math.min(left, right) + 1; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_112_Path_Sum/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._112_Path_Sum; 2 | 3 | public class Solution { 4 | 5 | /*给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有 6 | 节点值相加等于目标和。*/ 7 | 8 | public class TreeNode { 9 | int val; 10 | TreeNode left; 11 | TreeNode right; 12 | 13 | TreeNode(int x) { 14 | val = x; 15 | } 16 | } 17 | 18 | public boolean hasPathSum(TreeNode root, int sum) { 19 | if (root == null) { 20 | return false; 21 | } 22 | if (root.left == null && root.right == null && root.val == sum) { 23 | return true; 24 | } 25 | return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val); 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_121_Best_Time_to_Buy_and_Sell_Stock/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._121_Best_Time_to_Buy_and_Sell_Stock; 2 | 3 | public class Solution { 4 | 5 | /*给定一个数组，只能先买入后卖出，卖出价格要高于买入价格，求最大利润。只允许完成一笔交易*/ 6 | 7 | public int maxProfit(int[] prices) { 8 | int n = prices.length; 9 | if (n == 0) { 10 | return 0; 11 | } 12 | //最大利润 13 | int max = 0; 14 | //最小的价格 15 | int min = prices[0]; 16 | for (int i = 1; i < n; i++) { 17 | //比最小价格小就更新最小价格 18 | if (prices[i] < min) { 19 | min = prices[i]; 20 | } else { 21 | //比最小价格大就计算利润 22 | max = Math.max(max, prices[i] - min); 23 | } 24 | } 25 | return max; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_122_Best_Time_to_Buy_and_Sell_Stock_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._122_Best_Time_to_Buy_and_Sell_Stock_II; 2 | 3 | public class Solution { 4 | 5 | /*给定一个数组，要求先买入后卖出，可进行多次购买，求最大利润*/ 6 | 7 | public int maxProfit(int[] prices) { 8 | int total = 0; 9 | //后一个比前一个大的话直接累加 10 | for (int i = 1; i < prices.length; i++) { 11 | if (prices[i] > prices[i - 1]) { 12 | total += prices[i] - prices[i - 1]; 13 | } 14 | } 15 | return total; 16 | } 17 | } 18 | -------------------------------------------------------------------------------- /src/algorithm/_123_Best_Time_to_Buy_and_Sell_Stock_III/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._123_Best_Time_to_Buy_and_Sell_Stock_III; 2 | 3 | public class Solution { 4 | 5 | /*给定一个数组，它的第 i 个元素是一支给定的股票在第 i 天的价格。 6 | 设计一个算法来计算你所能获取的最大利润。你最多可以完成两笔交易。 7 | 注意: 你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。*/ 8 | 9 | 10 | /*采用状态机的思路，由于只能完成两笔交易，定义四个状态，分别是buy1,sell1,buy2,sell2*/ 11 | public int maxProfit(int[] prices) { 12 | int buy1 = Integer.MIN_VALUE, buy2 = Integer.MIN_VALUE; 13 | int sell1 = 0, sell2 = 0; 14 | for (int i = 0; i < prices.length; i++) { 15 | //买入价格更低的股 16 | buy1 = Math.max(buy1, -prices[i]); 17 | //卖出当前股或者不操作 18 | sell1 = Math.max(sell1, buy1 + prices[i]); 19 | //第二次买入，或者不操作 20 | buy2 = Math.max(buy2, sell1 - prices[i]); 21 | //第二次卖出，或者不操作 22 | sell2 = Math.max(sell2, buy2 + prices[i]); 23 | } 24 | return sell2; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_136_Single_Number/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._136_Single_Number; 2 | 3 | public class Solution { 4 | 5 | 6 | /*给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现两次。 7 | 找出那个只出现了一次的元素。*/ 8 | 9 | //0 ^ N = N N ^ N = 0 10 | 11 | /*N1 ^ N1 ^ N2 ^ N2 ^..............^ Nx ^ Nx ^ N 12 | = (N1^N1) ^ (N2^N2) ^..............^ (Nx^Nx) ^ N 13 | = 0 ^ 0 ^ ..........^ 0 ^ N 14 | = N*/ 15 | 16 | public int singleNumber(int[] nums) { 17 | int result = 0; 18 | for (int i = 0; i < nums.length; i++) { 19 | result ^= nums[i]; 20 | } 21 | return result; 22 | } 23 | } 24 | -------------------------------------------------------------------------------- /src/algorithm/_139_Word_Break/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._139_Word_Break; 2 | 3 | import java.util.List; 4 | 5 | public class Solution { 6 | 7 | /*给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict， 8 | 判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词。*/ 9 | 10 | //字典的单词没有使用次数限制这是一个完全背包的问题，该问题涉及到字典中单词的使用顺序，因此可理解为涉及顺序的完全背包问题。 11 | //和求解完全背包问题不同的地方在于求解顺序的完全背包问题时，对物品的迭代应该放在最里层。 12 | public boolean wordBreak(String s, List wordDict) { 13 | int length = s.length(); 14 | boolean[] dp = new boolean[length + 1]; 15 | dp[0] = true; 16 | for (int i = 1; i <= length; i++) { 17 | //物品迭代放里层 18 | for (String word : wordDict) { 19 | int len = word.length(); 20 | //判断是否单词是否匹配字符串 21 | if ((i >= len) && word.equals(s.substring(i - len, i))) { 22 | dp[i] = dp[i] || dp[i - len]; 23 | } 24 | } 25 | } 26 | return dp[length]; 27 | } 28 | } 29 | -------------------------------------------------------------------------------- /src/algorithm/_141_Linked_List_Cycle/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._141_Linked_List_Cycle; 2 | 3 | public class Solution { 4 | 5 | class ListNode { 6 | int val; 7 | ListNode next; 8 | 9 | ListNode(int x) { 10 | val = x; 11 | next = null; 12 | } 13 | } 14 | 15 | /*判断链表有无环*/ 16 | //定义两个指针，一个指针指向头指针，一次走一步，一个指针指向头指针下一个，一次走两步。如果指针能相等说明有环。如果为空则无环 17 | public boolean hasCycle(ListNode head) { 18 | if (head == null) return false; 19 | ListNode slow = head, fast = head; 20 | while (fast != null && fast.next != null) { 21 | fast = fast.next.next; 22 | slow = slow.next; 23 | if (fast == slow) { 24 | return true; 25 | } 26 | } 27 | return false; 28 | } 29 | 30 | 31 | } 32 | -------------------------------------------------------------------------------- /src/algorithm/_144_Binary_Tree_Preorder_Traversal/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._144_Binary_Tree_Preorder_Traversal; 2 | 3 | import java.util.ArrayList; 4 | import java.util.List; 5 | import java.util.Stack; 6 | 7 | 8 | public class Solution { 9 | 10 | class TreeNode { 11 | int val; 12 | TreeNode left; 13 | TreeNode right; 14 | 15 | TreeNode(int x) { 16 | val = x; 17 | } 18 | } 19 | 20 | 21 | /*非递归实现二叉树的前序遍历*/ 22 | public List preorderTraversal(TreeNode root) { 23 | List res = new ArrayList(); 24 | Stack stack = new Stack(); 25 | stack.push(root); 26 | while (!stack.empty()) { 27 | TreeNode node = stack.pop(); 28 | if (node == null) continue; 29 | res.add(node.val); 30 | stack.push(node.right); 31 | stack.push(node.left); 32 | } 33 | return res; 34 | } 35 | } 36 | -------------------------------------------------------------------------------- /src/algorithm/_145_Binary_Tree_Postorder_Traversal/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._145_Binary_Tree_Postorder_Traversal; 2 | 3 | 4 | import java.util.ArrayList; 5 | import java.util.Collections; 6 | import java.util.List; 7 | import java.util.Stack; 8 | 9 | public class Solution { 10 | 11 | class TreeNode { 12 | int val; 13 | TreeNode left; 14 | TreeNode right; 15 | 16 | TreeNode(int x) { 17 | val = x; 18 | } 19 | } 20 | 21 | 22 | /*非递归实现二叉树的后序遍历*/ 23 | public List postorderTraversal(TreeNode root) { 24 | List res = new ArrayList(); 25 | Stack stack = new Stack(); 26 | stack.push(root); 27 | while (!stack.empty()) { 28 | TreeNode node = stack.pop(); 29 | if (node == null) continue; 30 | res.add(node.val); 31 | stack.push(node.left); 32 | stack.push(node.right); 33 | } 34 | Collections.reverse(res); 35 | return res; 36 | } 37 | } 38 | -------------------------------------------------------------------------------- /src/algorithm/_152_Maximum_Product_Subarray/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._152_Maximum_Product_Subarray; 2 | 3 | public class Solution { 4 | 5 | 6 | /*给定一个整型数组，找出数组内连续子数组乘积的最大值*/ 7 | 8 | public int maxProduct(int[] nums) { 9 | if (nums == null || nums.length == 0) { 10 | return 0; 11 | } 12 | int max = nums[0], min = nums[0], result = nums[0]; 13 | //每一次都找出最大值和最小值，然后最大值和最小值分别乘以下一个数。因为负数乘负数为正数，所以要保存最小值 14 | for (int i = 1; i < nums.length; i++) { 15 | int temp = max; 16 | max = Math.max(Math.max(max * nums[i], min * nums[i]), nums[i]); 17 | min = Math.min(Math.min(temp * nums[i], min * nums[i]), nums[i]); 18 | if (max > result) { 19 | result = max; 20 | } 21 | } 22 | return result; 23 | } 24 | 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_154_Find_Minimum_in_Rotated_Sorted_Array_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._154_Find_Minimum_in_Rotated_Sorted_Array_II; 2 | 3 | public class Solution { 4 | 5 | /*假设按照升序排序的数组在预先未知的某个点上进行了旋转。 6 | ( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。 7 | 请找出其中最小的元素。 8 | 注意数组中可能存在重复的元素。*/ 9 | 10 | 11 | public int findMin(int[] nums) { 12 | int start = 0, end = nums.length - 1; 13 | while (start < end) { 14 | int mid = (start + end) / 2; 15 | if (nums[mid] > nums[end]) { 16 | start = mid + 1; 17 | } else if (nums[mid] < nums[end]) { 18 | end = mid; 19 | } else { 20 | //nums[mid]==nums[end],存在重复的元素，只需让end指针左移缩小范围 21 | end--; 22 | } 23 | } 24 | return nums[start]; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_160_Intersection_of_Two_Linked_Lists/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._160_Intersection_of_Two_Linked_Lists; 2 | 3 | public class Solution { 4 | 5 | public class ListNode { 6 | int val; 7 | ListNode next; 8 | ListNode(int x) { 9 | val = x; 10 | next = null; 11 | } 12 | } 13 | 14 | 15 | /*找到两个链表相交的结点，如果没有的话返回null*/ 16 | public ListNode getIntersectionNode(ListNode headA, ListNode headB) { 17 | ListNode nodeA = headA, nodeB = headB; 18 | //第二次迭代就能找出结点 19 | while (nodeA != nodeB) { 20 | nodeA = nodeA == null ? headB : nodeA.next; 21 | nodeB = nodeB == null ? headA : nodeB.next; 22 | } 23 | return nodeA; 24 | } 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_169_Majority_Element/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._169_Majority_Element; 2 | 3 | public class Solution { 4 | 5 | 6 | /*给定一个大小为 n 的数组，找到其中的众数。众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。*/ 7 | 8 | public int majorityElement(int[] num) { 9 | int major = num[0], count = 1; 10 | for (int i = 1; i < num.length; i++) { 11 | //如果count=0,重新在剩余的数字中找出众数 12 | if (count == 0) { 13 | count++; 14 | major = num[i]; 15 | } else if (major == num[i]) { 16 | count++; 17 | } else count--; 18 | 19 | } 20 | return major; 21 | } 22 | 23 | 24 | } 25 | -------------------------------------------------------------------------------- /src/algorithm/_190_Reverse_Bits/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._190_Reverse_Bits; 2 | 3 | public class Solution { 4 | 5 | /*颠倒给定的 32 位无符号整数的二进制位。*/ 6 | 7 | public int reverseBits(int n) { 8 | int result = 0; 9 | for (int i = 0; i < 32; i++) { 10 | result = result << 1; 11 | //n最右边位置的二进制位 12 | result |= (n & 1); 13 | //n右移一位 14 | n = n >> 1; 15 | } 16 | return result; 17 | } 18 | 19 | } 20 | -------------------------------------------------------------------------------- /src/algorithm/_198_House_Robber/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._198_House_Robber; 2 | 3 | public class Solution { 4 | 5 | /*题目描述：抢劫一排住户，但是不能抢邻近的住户，求最大抢劫量。 6 | 定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。 7 | 由于不能抢劫邻近住户，如果抢劫了第 i -1 个住户，那么就不能再抢劫第 i 个住户*/ 8 | 9 | 10 | //pre2表示当前房子数减2的最大值，pre1表示当前房子数减1的最大值，cur表示现在房子数的最大值 11 | public int rob(int[] nums) { 12 | int pre2 = 0, pre1 = 0; 13 | for (int i = 0; i < nums.length; i++) { 14 | int cur = Math.max(nums[i] + pre2, pre1); 15 | pre2 = pre1; 16 | pre1 = cur; 17 | } 18 | return pre1; 19 | } 20 | 21 | 22 | //采用数组来进行dp 23 | public int rob1(int[] nums) { 24 | if (nums.length == 0) return 0; 25 | int[] memo = new int[nums.length + 1]; 26 | memo[0] = 0; 27 | memo[1] = nums[0]; 28 | for (int i = 1; i < nums.length; i++) { 29 | int val = nums[i]; 30 | memo[i + 1] = Math.max(memo[i], memo[i - 1] + val); 31 | } 32 | return memo[nums.length]; 33 | } 34 | } 35 | -------------------------------------------------------------------------------- /src/algorithm/_199_Binary_Tree_Right_Side_View/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._199_Binary_Tree_Right_Side_View; 2 | 3 | import java.util.ArrayList; 4 | import java.util.List; 5 | 6 | public class Solution { 7 | 8 | public class TreeNode { 9 | int val; 10 | TreeNode left; 11 | TreeNode right; 12 | TreeNode(int x) { val = x; } 13 | } 14 | 15 | /*二叉树的右视图，返回从顶层到底层你看到的结点*/ 16 | 17 | 18 | public List rightSideView(TreeNode root) { 19 | List result = new ArrayList<>(); 20 | rightView(result, root, 0); 21 | return result; 22 | } 23 | 24 | public void rightView(List result, TreeNode node, int depth) { 25 | //如果为空，直接返回 26 | if (node == null) return; 27 | //只有满足大小和层数的等价关系才添加，一层只会添加一个结点 28 | if (result.size() == depth) { 29 | result.add(node.val); 30 | } 31 | //先递归右边，右边没有的才递归左边。左视图的话就相反 32 | rightView(result, node.right, depth + 1); 33 | rightView(result, node.left, depth + 1); 34 | } 35 | } 36 | -------------------------------------------------------------------------------- /src/algorithm/_202_Happy_Number/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._202_Happy_Number; 2 | 3 | public class Solution { 4 | 5 | /*判断一个数字是否是快乐数*/ 6 | 7 | /*一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和， 8 | 然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那 9 | 么这个数就是快乐数。*/ 10 | 11 | public boolean isHappy(int n) { 12 | //定义快指针和满指针 13 | int slow = n, fast = n; 14 | do { 15 | slow = digitSquareSum(slow); 16 | fast = digitSquareSum(fast); 17 | fast = digitSquareSum(fast); 18 | //如果能变为1表示是快乐数 19 | if (fast == 1) { 20 | return true; 21 | } 22 | } while (slow != fast); 23 | return false; 24 | } 25 | 26 | //计算平方和 27 | private int digitSquareSum(int n) { 28 | int sum = 0; 29 | while (n != 0) { 30 | int tmp = n % 10; 31 | sum += tmp * tmp; 32 | n /= 10; 33 | } 34 | return sum; 35 | } 36 | } 37 | -------------------------------------------------------------------------------- /src/algorithm/_205_Isomorphic_Strings/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._205_Isomorphic_Strings; 2 | 3 | public class Solution { 4 | 5 | /*给定两个字符串 s 和 t，判断它们是否是同构的。 6 | 如果 s 中的字符可以被替换得到 t ，那么这两个字符串是同构的。 7 | 所有出现的字符都必须用另一个字符替换，同时保留字符的顺序。 8 | 两个字符不能映射到同一个字符上，但字符可以映射自己本身。*/ 9 | 10 | 11 | //记录每个字符之前出现的索引位置，从而进行判断 12 | public boolean isIsomorphic(String s, String t) { 13 | int[] preIndexOfS = new int[256]; 14 | int[] preIndexOfT = new int[256]; 15 | for (int i = 0; i < s.length(); i++) { 16 | char sc = s.charAt(i), tc = t.charAt(i); 17 | //之前出现的位置不同，所以不是同构的 18 | if (preIndexOfS[sc] != preIndexOfT[tc]) { 19 | return false; 20 | } 21 | //记录出现的位置 22 | preIndexOfS[sc] = i + 1; 23 | preIndexOfT[tc] = i + 1; 24 | } 25 | return true; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_206_Reverse_Linked_List/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._206_Reverse_Linked_List; 2 | 3 | public class Solution { 4 | 5 | public class ListNode { 6 | int val; 7 | ListNode next; 8 | 9 | ListNode(int x) { 10 | val = x; 11 | } 12 | } 13 | 14 | 15 | /*反转一个单链表。*/ 16 | public ListNode reverseList(ListNode head) { 17 | ListNode pre = null, next = null; 18 | while (head != null) { 19 | next = head.next; 20 | head.next = pre; 21 | pre = head; 22 | head = next; 23 | } 24 | return pre; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_213_House_Robber2/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._213_House_Robber2; 2 | 3 | public class Solution { 4 | 5 | //由于房子是环形的，所以分两种情况考虑，一种是要偷第一间，一种是不偷第一间，然后取这两种情况的最大值， 6 | //这样的话这两种情况都能转化成在一条街上偷这些房子的最大值 7 | public int rob(int[] nums) { 8 | int length = nums.length; 9 | if (nums == null || length == 0) 10 | return 0; 11 | if (length == 1) 12 | return nums[0]; 13 | //抢劫第一间（也可能不抢）+不抢劫第一间 14 | return Math.max(rob(nums, 0, length - 2), rob(nums, 1, length - 1)); 15 | } 16 | 17 | public int rob(int[] nums, int index, int end) { 18 | int pre1 = 0, pre2 = 0; 19 | for (int i = index; i <= end; i++) { 20 | int cur = Math.max(nums[i] + pre2, pre1); 21 | pre2 = pre1; 22 | pre1 = cur; 23 | } 24 | return pre1; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_221_Maximal_Square/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._221_Maximal_Square; 2 | 3 | public class Solution { 4 | 5 | /*给定一个二维数组，找到最大的数字为1的正方形，返回面积*/ 6 | 7 | public int maximalSquare(char[][] matrix) { 8 | if (matrix.length == 0) { 9 | return 0; 10 | } 11 | int rows = matrix.length; 12 | int cols = matrix[0].length; 13 | int result = 0; 14 | //定义一个dp数组，数组的值dp[i][j]表示以原来二维数组的i-1，j-1这一格为右下角的正方形边长 15 | int[][] dp = new int[rows + 1][cols + 1]; 16 | for (int i = 1; i <= rows; i++) { 17 | for (int j = 1; j <= cols; j++) { 18 | //原来的数组有1的情况下才去更新值，否则就是0 19 | if (matrix[i - 1][j - 1] == '1') { 20 | dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i - 1][j - 1], dp[i][j - 1])) + 1; 21 | //更新最大正方形边长 22 | result = Math.max(result, dp[i][j]); 23 | } 24 | } 25 | } 26 | //面积 27 | return result * result; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_222_Count_Complete_Tree_Nodes/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._222_Count_Complete_Tree_Nodes; 2 | 3 | public class Solution { 4 | //todo 5 | } 6 | -------------------------------------------------------------------------------- /src/algorithm/_226_Invert_Binary_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._226_Invert_Binary_Tree; 2 | 3 | public class Solution { 4 | 5 | public class TreeNode { 6 | int val; 7 | TreeNode left; 8 | TreeNode right; 9 | 10 | TreeNode(int x) { 11 | val = x; 12 | } 13 | } 14 | 15 | 16 | /*反转二叉树*/ 17 | public TreeNode invertTree(TreeNode root) { 18 | if (root != null) { 19 | TreeNode temp = root.left; 20 | root.left = root.right; 21 | root.right = temp; 22 | invertTree(root.left); 23 | invertTree(root.right); 24 | } 25 | return root; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_230_Kth_Smallest_Element_in_a_BST/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._230_Kth_Smallest_Element_in_a_BST; 2 | 3 | 4 | public class Solution { 5 | 6 | public class TreeNode { 7 | int val; 8 | TreeNode left; 9 | TreeNode right; 10 | 11 | TreeNode(int x) { 12 | val = x; 13 | } 14 | } 15 | 16 | 17 | /*给定一个二叉搜索树，编写一个函数 kthSmallest 来查找其中第 k 个最小的元素。*/ 18 | 19 | private int count = 0; 20 | private int number = 0; 21 | 22 | public int kthSmallest(TreeNode root, int k) { 23 | count = k; 24 | helper(root); 25 | return number; 26 | } 27 | 28 | private void helper(TreeNode root){ 29 | if(root.left != null){ 30 | helper(root.left); 31 | } 32 | count--; 33 | if(count == 0){ 34 | number = root.val; 35 | } 36 | if(root.right != null){ 37 | helper(root.right); 38 | } 39 | } 40 | } 41 | -------------------------------------------------------------------------------- /src/algorithm/_235_Lowest_Common_Ancestor_of_a_Binary_Search_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._235_Lowest_Common_Ancestor_of_a_Binary_Search_Tree; 2 | 3 | public class Solution { 4 | 5 | class TreeNode { 6 | int val; 7 | TreeNode left; 8 | TreeNode right; 9 | 10 | public TreeNode(int val) { 11 | this.val = val; 12 | } 13 | } 14 | 15 | /*二叉查找树的两个结点的最近公共祖先*/ 16 | 17 | public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { 18 | //如果两个结点都在左边的情况下 19 | if (root.val > p.val && root.val > q.val) { 20 | return lowestCommonAncestor(root.left, p, q); 21 | } 22 | //如果两个结点都在右边的情况下 23 | if (root.val < p.val && root.val < q.val) { 24 | return lowestCommonAncestor(root.right, p, q); 25 | } 26 | //两个结点一左一右或者其中一个是祖先结点的情况下直接返回祖先结点 27 | return root; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_236_Lowest_Common_Ancestor_of_a_Binary_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._236_Lowest_Common_Ancestor_of_a_Binary_Tree; 2 | 3 | public class Solution { 4 | 5 | class TreeNode { 6 | int val; 7 | TreeNode left; 8 | TreeNode right; 9 | 10 | TreeNode(int x) { 11 | val = x; 12 | } 13 | } 14 | 15 | /*普通二叉树的最近公共祖先*/ 16 | //后序遍历二叉树，左子树和右子树分别搜索 17 | public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { 18 | //终止条件 19 | if (root == null || root == p || root == q) { 20 | return root; 21 | } 22 | //左子树查找 23 | TreeNode left = lowestCommonAncestor(root.left, p, q); 24 | //右子树查找 25 | TreeNode right = lowestCommonAncestor(root.right, p, q); 26 | //如果左子树和右子树都找到了，就返回根节点 27 | if (left != null && right != null) { 28 | return root; 29 | } 30 | //如果左右子树只有一个找到，那就返回找到的。如果都没有找到，那就返回空 31 | return left != null ? left : right; 32 | } 33 | } 34 | -------------------------------------------------------------------------------- /src/algorithm/_238_Product_of_Array_Except_Self/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._238_Product_of_Array_Except_Self; 2 | 3 | public class Solution { 4 | 5 | /*给定一个整数数组，输出除去该位置元素的其他元素的乘积的数组,时间复杂度要求O(n)*/ 6 | /* 7 | Numbers: 2 3 4 5 8 | Lefts: 1 2 2*3 2*3*4 9 | Rights: 3*4*5 4*5 5 1 10 | */ 11 | 12 | public int[] productExceptSelf(int[] nums) { 13 | int n = nums.length; 14 | int[] result = new int[n]; 15 | //确定左边方向的成绩 16 | int left = 1; 17 | for (int i = 0; i < n; i++) { 18 | if (i > 0) { 19 | left = left * nums[i - 1]; 20 | } 21 | result[i] = left; 22 | } 23 | //确定右边方向的成绩 24 | int right = 1; 25 | for (int i = n - 1; i >= 0; i--) { 26 | if (i < n - 1) { 27 | right = right * nums[i + 1]; 28 | } 29 | //左边方向乘以右边方向 30 | result[i] *= right; 31 | } 32 | return result; 33 | } 34 | } 35 | -------------------------------------------------------------------------------- /src/algorithm/_239_Sliding_Window_Maximum/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._239_Sliding_Window_Maximum; 2 | 3 | import java.util.PriorityQueue; 4 | 5 | public class Solution { 6 | 7 | 8 | /*给定一个数组 nums，有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。 9 | 你只可以看到在滑动窗口内的 k 个数字。滑动窗口每次只向右移动一位。*/ 10 | 11 | public int[] maxSlidingWindow(int[] nums, int k) { 12 | if (nums.length < k || k < 1) { 13 | return new int[0]; 14 | } 15 | int[] result = new int[nums.length - k + 1]; 16 | //大顶堆 17 | PriorityQueue heap = new PriorityQueue<>((o1, o2) -> o2 - o1); 18 | for (int i = 0; i < k; i++) { 19 | heap.add(nums[i]); 20 | } 21 | result[0] = heap.peek(); 22 | //滑动窗口，记录大顶堆的堆顶值 23 | for (int i = 0, j = k; j < nums.length; i++, j++) { 24 | heap.remove(nums[i]); 25 | heap.add(nums[j]); 26 | result[i + 1] = heap.peek(); 27 | } 28 | return result; 29 | } 30 | } 31 | -------------------------------------------------------------------------------- /src/algorithm/_240_Search_a_2D_Matrix_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._240_Search_a_2D_Matrix_II; 2 | 3 | public class Solution { 4 | 5 | 6 | /*编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性： 7 | 每行的元素从左到右升序排列。 8 | 每列的元素从上到下升序排列。*/ 9 | 10 | 11 | //从左下角位置开始查找 12 | public boolean searchMatrix(int[][] matrix, int target) { 13 | if (matrix == null || matrix.length == 0) { 14 | return false; 15 | } 16 | int row = matrix.length - 1; 17 | int col = 0; 18 | while (col < matrix[0].length && row >= 0) { 19 | if (matrix[row][col] < target) { 20 | col++; 21 | } else if (matrix[row][col] > target) { 22 | row--; 23 | } else { 24 | return true; 25 | } 26 | } 27 | return false; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_242_Valid_Anagram/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._242_Valid_Anagram; 2 | 3 | public class Solution { 4 | 5 | /*给定两个字符串 s 和 t ，编写一个函数来判断 t 是否是 s 的字母异位词。*/ 6 | 7 | //只需要判断s中出现的字符，t是否全部包含即可 8 | public boolean isAnagram(String s, String t) { 9 | int[] nums = new int[26]; 10 | for (int i = 0; i < s.length(); i++) nums[s.charAt(i) - 'a']++; 11 | for (int i = 0; i < t.length(); i++) nums[t.charAt(i) - 'a']--; 12 | for (int num : nums) if (num != 0) return false; 13 | return true; 14 | } 15 | } 16 | -------------------------------------------------------------------------------- /src/algorithm/_260_Single_Number_III/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._260_Single_Number_III; 2 | 3 | public class Solution { 4 | 5 | /*给定一个整数数组 nums，其中恰好有两个元素只出现一次，其余所有元素均出现两次。 6 | 找出只出现一次的那两个元素。*/ 7 | 8 | public int[] singleNumber(int[] nums) { 9 | int diff = 0; 10 | //所有异或，最后得出的是两个不重复的元素异或的结果 11 | for (int num : nums) { 12 | diff ^= num; 13 | } 14 | //找到最右边为1的数字，利用这个数字区分两个元素 15 | diff &= -diff; 16 | int[] result = new int[2]; 17 | for (int num : nums) { 18 | if ((num & diff) == diff) { 19 | result[0] ^= num; 20 | } else { 21 | result[1] ^= num; 22 | } 23 | } 24 | return result; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_263_Ugly_Number/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._263_Ugly_Number; 2 | 3 | public class Solution { 4 | 5 | /*编写一个程序判断给定的数是否为丑数。丑数就是只包含质因数 2, 3, 5 的正整数。*/ 6 | 7 | public boolean isUgly(int num) { 8 | for (int i = 2; i < 6 && num > 0; i++) { 9 | while (num % i == 0) { 10 | num /= i; 11 | } 12 | } 13 | return num == 1; 14 | } 15 | } 16 | -------------------------------------------------------------------------------- /src/algorithm/_264_Ugly_Number_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._264_Ugly_Number_II; 2 | 3 | public class Solution { 4 | 5 | /*编写一个程序，找出第 n 个丑数。丑数就是只包含质因数 2, 3, 5 的正整数。*/ 6 | 7 | public int nthUglyNumber(int n) { 8 | if (n < 6) { 9 | return n; 10 | } 11 | int[] nums = new int[n]; 12 | nums[0] = 1; 13 | int i2 = 0, i3 = 0, i5 = 0; 14 | for (int i = 1; i < n; i++) { 15 | int next2 = nums[i2] * 2, next3 = nums[i3] * 3, next5 = nums[i5] * 5; 16 | //最小值就是下一个丑数 17 | nums[i] = Math.min(next2, Math.min(next3, next5)); 18 | if (nums[i] == next2) i2++; 19 | if (nums[i] == next3) i3++; 20 | if (nums[i] == next5) i5++; 21 | } 22 | return nums[n - 1]; 23 | } 24 | } 25 | -------------------------------------------------------------------------------- /src/algorithm/_268_Missing_Number/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._268_Missing_Number; 2 | 3 | public class Solution { 4 | 5 | /*给定一个包含 0, 1, 2, ..., n 中 n 个数的序列，找出 0 .. n 中没有出现在序列中的那个数。*/ 6 | 7 | public int missingNumber(int[] nums) { 8 | if (nums == null || nums.length == 0) { 9 | return 0; 10 | } 11 | int x = 0, i = 0; 12 | for (; i < nums.length; i++) { 13 | x = x ^ i ^ nums[i]; 14 | } 15 | return x ^ i; 16 | } 17 | 18 | 19 | } 20 | -------------------------------------------------------------------------------- /src/algorithm/_279_Perfect_Squares/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._279_Perfect_Squares; 2 | 3 | import java.util.Arrays; 4 | 5 | public class Solution { 6 | 7 | /*给定一个正整数，找到最少能构成该正整数的完全平方数的个数*/ 8 | 9 | public int numSquares(int n) { 10 | int[] dp = new int[n + 1]; 11 | Arrays.fill(dp, Integer.MAX_VALUE); 12 | //初始状态 13 | dp[0] = 0; 14 | //每一次循环确定i这个正整数的最少完全平方数个数 15 | for (int i = 1; i <= n; i++) { 16 | //尝试不同的完全平方数，找出最小值 17 | for (int j = 1; i - j * j >= 0; j++) { 18 | dp[i] = Math.min(dp[i - j * j] + 1, dp[i]); 19 | } 20 | } 21 | return dp[n]; 22 | } 23 | } 24 | -------------------------------------------------------------------------------- /src/algorithm/_283_Move_Zeroes/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._283_Move_Zeroes; 2 | 3 | public class Solution { 4 | 5 | /*给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。*/ 6 | 7 | public void moveZeroes(int[] nums) { 8 | if(nums == null || nums.length == 0){ 9 | return ; 10 | } 11 | int index = 0; 12 | for(int num : nums){ 13 | if(num != 0){ 14 | nums[index++] = num; 15 | } 16 | } 17 | while(index < nums.length){ 18 | nums[index++] = 0; 19 | } 20 | } 21 | } 22 | -------------------------------------------------------------------------------- /src/algorithm/_300_Longest_Increasing_Subsequence/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._300_Longest_Increasing_Subsequence; 2 | 3 | public class Solution { 4 | 5 | /*求数组的最长递增子序列，子序列可以不连续的*/ 6 | 7 | //使用动态规划求解，dp[n]表示以n结尾的最长递增子序列，时间复杂度O（n2） 8 | public int lengthOfLIS(int[] nums) { 9 | int length = nums.length; 10 | int[] dp = new int[length]; 11 | for (int i = 0; i < length; i++) { 12 | //初始长度 13 | int max = 1; 14 | //循环后可以确定以nums[i]结尾的最长递增子序列 15 | for (int j = 0; j < i; j++) { 16 | //如果nums[j]> 1] + (i & 1); 15 | } 16 | return result; 17 | } 18 | } 19 | -------------------------------------------------------------------------------- /src/algorithm/_343_Integer_Break/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._343_Integer_Break; 2 | 3 | public class Solution { 4 | 5 | //分割整数为至少两个正数，算出分割后的数的最大乘积 6 | 7 | public int integerBreak(int n) { 8 | int[] dp = new int[n + 1]; 9 | dp[1] = 1; 10 | for (int i = 2; i <= n; i++) { 11 | for (int j = i - 1; j > 0; j--) { 12 | //从这三种情况中取最大值 13 | dp[i] = Math.max(dp[i], Math.max(dp[i - j] * j, j * (i - j))); 14 | } 15 | } 16 | return dp[n]; 17 | } 18 | } 19 | -------------------------------------------------------------------------------- /src/algorithm/_344_Reverse_String/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._344_Reverse_String; 2 | 3 | public class Solution { 4 | 5 | /*编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 char[] 的形式给出。*/ 6 | 7 | public void reverseString(char[] s) { 8 | int start = 0, end = s.length - 1; 9 | while (start < end) { 10 | char temp = s[start]; 11 | s[start] = s[end]; 12 | s[end] = temp; 13 | start++; 14 | end--; 15 | } 16 | } 17 | } 18 | -------------------------------------------------------------------------------- /src/algorithm/_376_Wiggle_Subsequence/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._376_Wiggle_Subsequence; 2 | 3 | public class Solution { 4 | 5 | /*如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差 6 | （如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。 7 | 例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。 8 | 相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值 9 | 都是正数，第二个序列是因为它的最后一个差值为零。 10 | 给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些 11 | （也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。*/ 12 | 13 | 14 | public int wiggleMaxLength(int[] nums) { 15 | if (nums == null || nums.length == 0) { 16 | return 0; 17 | } 18 | //初始状态 19 | int up = 1, down = 1; 20 | for (int i = 1; i < nums.length; i++) { 21 | if (nums[i] > nums[i - 1]) { 22 | //状态转移 23 | up = down + 1; 24 | } else if (nums[i] < nums[i - 1]) { 25 | //状态转移 26 | down = up + 1; 27 | } 28 | } 29 | return Math.max(up, down); 30 | } 31 | 32 | } 33 | -------------------------------------------------------------------------------- /src/algorithm/_377_Combination_Sum_IV/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._377_Combination_Sum_IV; 2 | 3 | import java.util.Arrays; 4 | 5 | public class Solution { 6 | 7 | /*数组中的数字构成目标数的组合数，数字可重复*/ 8 | 9 | public int combinationSum4(int[] nums, int target) { 10 | if (nums == null || nums.length == 0) { 11 | return 0; 12 | } 13 | //数组中的每一个数表示该数字的组合数 14 | int[] dp = new int[target + 1]; 15 | //初始化状态 16 | dp[0] = 1; 17 | //对数组进行排序 18 | Arrays.sort(nums); 19 | //计算目标数字的组合数 20 | for (int i = 1; i <= target; i++) { 21 | //当目标数字分别减去数组中的数字的组合数相加 22 | for (int j = 0; j < nums.length && nums[j] <= i; j++) { 23 | //状态转移方程 24 | dp[i] += dp[i - nums[j]]; 25 | } 26 | } 27 | return dp[target]; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_387_First_Unique_Character_in_a_String/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._387_First_Unique_Character_in_a_String; 2 | 3 | public class Solution { 4 | 5 | /*给定一个字符串，找到它的第一个不重复的字符，并返回它的索引。如果不存在，则返回 -1。*/ 6 | 7 | public int firstUniqChar(String s) { 8 | int[] freq = new int[26]; 9 | //存放在数组中 10 | for (int i = 0; i < s.length(); i++) { 11 | char c = s.charAt(i); 12 | freq[c - 'a']++; 13 | } 14 | for (int i = 0; i < s.length(); i++) { 15 | char c = s.charAt(i); 16 | if (freq[c - 'a'] == 1) { 17 | return i; 18 | } 19 | } 20 | return -1; 21 | } 22 | } 23 | -------------------------------------------------------------------------------- /src/algorithm/_392_Is_Subsequence/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._392_Is_Subsequence; 2 | 3 | public class Solution { 4 | 5 | //判断s是否是t的子序列 6 | public boolean isSubsequence(String s, String t) { 7 | int index = -1; 8 | for (char c : s.toCharArray()) { 9 | //indexOf代表从index+1开始查找c元素，返回对应索引 10 | index = t.indexOf(c, index + 1); 11 | if (index == -1) { 12 | return false; 13 | } 14 | } 15 | return true; 16 | 17 | } 18 | } 19 | -------------------------------------------------------------------------------- /src/algorithm/_404_Sum_of_Left_Leaves/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._404_Sum_of_Left_Leaves; 2 | 3 | public class Solution { 4 | 5 | /*计算给定二叉树的所有左叶子之和。*/ 6 | 7 | public class TreeNode { 8 | int val; 9 | TreeNode left; 10 | TreeNode right; 11 | 12 | TreeNode(int x) { 13 | val = x; 14 | } 15 | } 16 | 17 | public int sumOfLeftLeaves(TreeNode root) { 18 | if (root == null) { 19 | return 0; 20 | } 21 | //左结点是叶子节点 22 | if (isLeaf(root.left)) { 23 | return root.left.val + sumOfLeftLeaves(root.right); 24 | } 25 | 26 | return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right); 27 | } 28 | 29 | //是否是叶子节点 30 | private boolean isLeaf(TreeNode node) { 31 | if (node == null) { 32 | return false; 33 | } 34 | return node.left == null && node.right == null; 35 | } 36 | } 37 | -------------------------------------------------------------------------------- /src/algorithm/_405_Convert_a_Number_to_Hexadecimal/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._405_Convert_a_Number_to_Hexadecimal; 2 | 3 | public class Solution { 4 | 5 | /*给定一个整数，编写一个算法将这个数转换为十六进制数。对于负整数，我们通常使用补码运算方法。*/ 6 | 7 | public String toHex(int num) { 8 | if (num == 0) { 9 | return "0"; 10 | } 11 | char[] map = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'}; 12 | StringBuilder stb = new StringBuilder(); 13 | while (num != 0) { 14 | //0b1111表示1111是个二进制 15 | stb.append(map[num & 0b1111]); 16 | //因为考虑的是补码形式，因此符号位就不能有特殊的意义，所以无符号右移四位 17 | num >>>= 4; 18 | } 19 | return stb.reverse().toString(); 20 | } 21 | } 22 | -------------------------------------------------------------------------------- /src/algorithm/_409_Longest_Palindrome/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._409_Longest_Palindrome; 2 | 3 | public class Solution { 4 | 5 | /*给定一个包含大写字母和小写字母的字符串，找到通过这些字母构造成的最长的回文串。 6 | 在构造过程中，请注意区分大小写。比如 "Aa" 不能当做一个回文字符串。*/ 7 | 8 | 9 | public int longestPalindrome(String s) { 10 | int[] cnts = new int[256]; 11 | //记录每个字符出现的次数 12 | for (char c : s.toCharArray()) { 13 | cnts[c]++; 14 | } 15 | int palindrome = 0; 16 | for (int cnt : cnts) { 17 | //能构成回文的字符数量 18 | palindrome += (cnt / 2) * 2; 19 | } 20 | //小于字符串总长度的话，还能增加单个字符放在回文串正中间 21 | if (palindrome < s.length()) { 22 | palindrome++; 23 | } 24 | return palindrome; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_413_Arithmetic_Slices/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._413_Arithmetic_Slices; 2 | 3 | public class Solution { 4 | 5 | /*如果一个数列至少有三个元素，并且任意两个相邻元素之差相同，则称该数列为等差数列。 6 | 函数要返回数组 A 中所有为等差数组的子数组个数。*/ 7 | 8 | 9 | //dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。 10 | public int numberOfArithmeticSlices(int[] A) { 11 | if (A == null || A.length == 0) { 12 | return 0; 13 | } 14 | int[] dp = new int[A.length]; 15 | for (int i = 2; i < A.length; i++) { 16 | if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) { 17 | dp[i] = dp[i - 1] + 1; 18 | } 19 | } 20 | //因为递增子区间不一定以最后一个元素为结尾，可以是任意一个元素结尾，因此需要返回 dp 数组累加的结果。 21 | int result = 0; 22 | for (int d : dp) { 23 | result += d; 24 | } 25 | return result; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_448_Find_All_Numbers_Disappeared_in_an_Array/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._448_Find_All_Numbers_Disappeared_in_an_Array; 2 | 3 | import java.util.ArrayList; 4 | import java.util.List; 5 | 6 | public class Solution { 7 | 8 | /*给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次， 9 | 另一些只出现一次。找到所有在 [1, n] 范围之间没有出现在数组中的数字。时间复杂度o(n)*/ 10 | 11 | 12 | public List findDisappearedNumbers(int[] nums) { 13 | List list = new ArrayList<>(); 14 | for(int i = 0;i < nums.length;i++){ 15 | int val = Math.abs(nums[i]) - 1; 16 | //标记该位置的数为负数 17 | if(nums[val] > 0){ 18 | nums[val] = -nums[val]; 19 | } 20 | } 21 | for(int i = 0;i < nums.length;i++){ 22 | //如果为正数，表示该位置没有被标记过，也就表面该数组没有出现过这个数字 23 | if(nums[i] > 0){ 24 | list.add(i+1); 25 | } 26 | } 27 | return list; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_455_Assign_Cookies/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._455_Assign_Cookies; 2 | 3 | import java.util.Arrays; 4 | 5 | public class Solution { 6 | 7 | /*题目描述：每个孩子都有一个满足度，每个饼干都有一个大小，只有饼干的大小大于等于一个孩子的满足度，该孩子才会获得满足。求解最多可以获得满足的孩子数量。 8 | 9 | 给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因为最小的孩子最容易得到满足，所以先满足最小的孩子。*/ 10 | public int findContentChildren(int[] g, int[] s) { 11 | Arrays.sort(g); 12 | Arrays.sort(s); 13 | int gi = 0, si = 0; 14 | while (gi < g.length && si < s.length) { 15 | if (g[gi] <= s[si]) { 16 | gi++; 17 | } 18 | si++; 19 | } 20 | return gi; 21 | } 22 | 23 | 24 | } 25 | -------------------------------------------------------------------------------- /src/algorithm/_461_Hamming_Distance/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._461_Hamming_Distance; 2 | 3 | public class Solution { 4 | 5 | /*两个整数之间的汉明距离指的是这两个数字对应二进制位不同的位置的数目。 6 | 给出两个整数 x 和 y，计算它们之间的汉明距离。*/ 7 | 8 | //两数异或，二进制不同的位置的值为1，只需计算有多少个1即可 9 | public int hammingDistance(int x, int y) { 10 | int z = x ^ y; 11 | int count = 0; 12 | while (z != 0) { 13 | if ((z & 1) == 1) { 14 | count++; 15 | } 16 | z = z >> 1; 17 | } 18 | return count; 19 | } 20 | } 21 | -------------------------------------------------------------------------------- /src/algorithm/_462_Minimum_Moves_to_Equal_Array_Elements_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._462_Minimum_Moves_to_Equal_Array_Elements_II; 2 | 3 | import java.util.Arrays; 4 | 5 | public class Solution { 6 | 7 | /*给定一个非空整数数组，找到使所有数组元素相等所需的最小移动数，其中每次移动可将选定的一个元素 8 | 加1或减1。您可以假设数组的长度最多为10000。*/ 9 | 10 | //a,b表示中间值两边的两个数，向中间移动的次数等于b-a 11 | public int minMoves2(int[] nums) { 12 | Arrays.sort(nums); 13 | int start = 0, end = nums.length - 1; 14 | int move = 0; 15 | while (start <= end) { 16 | move += nums[end] - nums[start]; 17 | start++; 18 | end--; 19 | } 20 | return move; 21 | } 22 | } 23 | -------------------------------------------------------------------------------- /src/algorithm/_485_Max_Consecutive_Ones/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._485_Max_Consecutive_Ones; 2 | 3 | public class Solution { 4 | 5 | /*给定一个二进制数组， 计算其中最大连续1的个数。*/ 6 | 7 | public int findMaxConsecutiveOnes(int[] nums) { 8 | int max = 0, cur = 0; 9 | for (int num : nums) { 10 | cur = num == 0 ? 0 : cur + 1; 11 | max = Math.max(max, cur); 12 | } 13 | return max; 14 | } 15 | } 16 | -------------------------------------------------------------------------------- /src/algorithm/_496_Next_Greater_Element_I/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._496_Next_Greater_Element_I; 2 | 3 | import java.util.HashMap; 4 | import java.util.Map; 5 | import java.util.Stack; 6 | 7 | public class Solution { 8 | 9 | /*给定两个没有重复元素的数组 nums1 和 nums2 ，其中nums1 是 nums2 的子集。 10 | 找到 nums1 中每个元素在 nums2 中的下一个比其大的值。 11 | nums1 中数字 x 的下一个更大元素是指 x 在 nums2 中对应位置的右边的第一个比 x 大的元素。 12 | 如果不存在，对应位置输出-1。*/ 13 | 14 | 15 | 16 | public int[] nextGreaterElement(int[] nums1, int[] nums2) { 17 | //用于保存每个元素的下一个比其大的值 18 | Map map = new HashMap<>(); 19 | Stack stack = new Stack<>(); 20 | for (int num : nums2) { 21 | //栈不为空而且找到比栈顶元素大的值 22 | while (!stack.empty() && stack.peek() < num) { 23 | map.put(stack.pop(), num); 24 | } 25 | stack.push(num); 26 | } 27 | for (int i = 0; i < nums1.length; i++) { 28 | nums1[i] = map.getOrDefault(nums1[i], -1); 29 | } 30 | return nums1; 31 | } 32 | } 33 | -------------------------------------------------------------------------------- /src/algorithm/_503_Next_Greater_Element_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._503_Next_Greater_Element_II; 2 | 3 | import java.util.Arrays; 4 | import java.util.Stack; 5 | 6 | public class Solution { 7 | 8 | /*给定一个循环数组（最后一个元素的下一个元素是数组的第一个元素），输出每个元素的 9 | 下一个更大元素。数字 x 的下一个更大的元素是按数组遍历顺序，这个数字之后的第一个 10 | 比它更大的数，这意味着你应该循环地搜索它的下一个更大的数。如果不存在，则输出 -1。*/ 11 | 12 | 13 | public int[] nextGreaterElements(int[] nums) { 14 | int n = nums.length; 15 | int[] next = new int[n]; 16 | Arrays.fill(next, -1); 17 | //栈保存之前的元素 18 | Stack stack = new Stack<>(); 19 | //2*n表示要遍历两遍 20 | for (int i = 0; i < 2 * n; i++) { 21 | int num = nums[i % n]; 22 | //栈不为空而且栈顶索引指向的元素比当前元素小，更新next数组的对应索引元素为当前元素 23 | while (!stack.empty() && nums[stack.peek()] < num) { 24 | next[stack.pop()] = num; 25 | } 26 | if (i < n) { 27 | stack.push(i); 28 | } 29 | } 30 | return next; 31 | } 32 | } 33 | -------------------------------------------------------------------------------- /src/algorithm/_504_Base_7/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._504_Base_7; 2 | 3 | public class Solution { 4 | 5 | /*给定一个整数，将其转化为7进制，并以字符串形式输出。*/ 6 | 7 | public String convertToBase7(int num) { 8 | if (num == 0) { 9 | return "0"; 10 | } 11 | StringBuilder stb = new StringBuilder(); 12 | boolean isNegative = num < 0; 13 | if (isNegative) { 14 | num = -num; 15 | } 16 | while (num > 0) { 17 | //取余数 18 | stb.append(num % 7); 19 | num /= 7; 20 | } 21 | //字符串反转 22 | String res = stb.reverse().toString(); 23 | return isNegative ? "-" + res : res; 24 | } 25 | } 26 | -------------------------------------------------------------------------------- /src/algorithm/_513_Find_Bottom_Left_Tree_Value/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._513_Find_Bottom_Left_Tree_Value; 2 | 3 | import java.util.LinkedList; 4 | import java.util.Queue; 5 | 6 | public class Solution { 7 | 8 | 9 | public class TreeNode { 10 | int val; 11 | TreeNode left; 12 | TreeNode right; 13 | 14 | TreeNode(int x) { 15 | val = x; 16 | } 17 | } 18 | 19 | /*给定一个二叉树，找到最下面一行的最左边的结点*/ 20 | 21 | public int findBottomLeftValue(TreeNode root) { 22 | if (root == null) return 0; 23 | Queue queue = new LinkedList<>(); 24 | queue.add(root); 25 | while (!queue.isEmpty()) { 26 | root = queue.poll(); 27 | if (root.right != null) { 28 | queue.add(root.right); 29 | } 30 | if (root.left != null) { 31 | queue.add(root.left); 32 | } 33 | } 34 | return root.val; 35 | } 36 | } 37 | -------------------------------------------------------------------------------- /src/algorithm/_518_Coin_Change_2/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._518_Coin_Change_2; 2 | 3 | public class Solution { 4 | 5 | /*给定不同面额的硬币和一个总金额。写出函数来计算可以凑成总金额的硬币组合数。 6 | 假设每一种面额的硬币有无限个。*/ 7 | 8 | 9 | //完全背包问题 10 | public int change(int amount, int[] coins) { 11 | int[] dp = new int[amount + 1]; 12 | dp[0] = 1; 13 | for (int coin : coins) { 14 | for (int i = coin; i <= amount; i++) {//完全背包，硬币可重复，采用正序 15 | dp[i] = dp[i] + dp[i - coin]; 16 | } 17 | } 18 | return dp[amount]; 19 | } 20 | 21 | } 22 | -------------------------------------------------------------------------------- /src/algorithm/_530_Minimum_Absolute_Difference_in_BST/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._530_Minimum_Absolute_Difference_in_BST; 2 | 3 | public class Solution { 4 | 5 | /*给定一个所有节点为非负值的二叉搜索树，求树中任意两节点的差的绝对值的最小值。*/ 6 | 7 | public class TreeNode { 8 | int val; 9 | TreeNode left; 10 | TreeNode right; 11 | 12 | TreeNode(int x) { 13 | val = x; 14 | } 15 | } 16 | 17 | 18 | private int min = Integer.MAX_VALUE; 19 | private TreeNode pre = null; 20 | 21 | public int getMinimumDifference(TreeNode root) { 22 | inOrder(root); 23 | return min; 24 | } 25 | 26 | //中序遍历 27 | private void inOrder(TreeNode root) { 28 | if (root == null) { 29 | return; 30 | } 31 | inOrder(root.left); 32 | //当前节点和前一个节点的差值 33 | if (pre != null) { 34 | min = Math.min(min, root.val - pre.val); 35 | } 36 | //更新前一个结点 37 | pre = root; 38 | inOrder(root.right); 39 | } 40 | } 41 | -------------------------------------------------------------------------------- /src/algorithm/_538_Convert_BST_to_Greater_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._538_Convert_BST_to_Greater_Tree; 2 | 3 | public class Solution { 4 | 5 | /*给定一个二叉搜索树（Binary Search Tree），把它转换成为累加树（Greater Tree)， 6 | 使得每个节点的值是原来的节点值加上所有大于它的节点值之和。*/ 7 | 8 | 9 | public class TreeNode { 10 | int val; 11 | TreeNode left; 12 | TreeNode right; 13 | TreeNode(int x) { val = x; } 14 | } 15 | 16 | 17 | //全局变量 18 | int sum = 0; 19 | 20 | public TreeNode convertBST(TreeNode root) { 21 | convert(root); 22 | return root; 23 | } 24 | 25 | //右根左，中序遍历反过来 26 | /*一直递归右结点，找到最右的结点，更新它的值为结点值加上累加值，并将累加值更新为该结点的值。再递归左结点*/ 27 | private void convert(TreeNode cur){ 28 | if(cur == null){ 29 | return; 30 | } 31 | convert(cur.right); 32 | cur.val += sum; 33 | sum = cur.val; 34 | convert(cur.left); 35 | } 36 | } 37 | -------------------------------------------------------------------------------- /src/algorithm/_540_Single_Element_in_a_Sorted_Array/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._540_Single_Element_in_a_Sorted_Array; 2 | 3 | public class Solution { 4 | 5 | /*给定一个只包含整数的有序数组，每个元素都会出现两次，唯有一个数只会出现一次，找出这个数。 6 | 要求O(log n)时间复杂度*/ 7 | 8 | public int singleNonDuplicate(int[] nums) { 9 | int L = 0, R = nums.length - 1; 10 | while (L < R) { 11 | //中位数位置 12 | int mid = L + (R - L) / 2; 13 | //如果中位数的位置是奇数，那么进行减1操作，确保左边区间和右边区间的长度都为偶数 14 | if (mid % 2 == 1) { 15 | mid--; 16 | } 17 | //相等表示右边区间除去nums[mid + 1]之后，长度为奇数，那么该数肯定在这里面 18 | if (nums[mid] == nums[mid + 1]) { 19 | L = mid + 2; 20 | //反之表示右边区间的偶数都是两两匹配，该数不在这里面。 21 | } else { 22 | R = mid; 23 | } 24 | } 25 | return nums[L]; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_543_Diameter_of_Binary_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._543_Diameter_of_Binary_Tree; 2 | 3 | public class Solution { 4 | 5 | public class TreeNode { 6 | int val; 7 | TreeNode left; 8 | TreeNode right; 9 | TreeNode(int x) { val = x; } 10 | } 11 | 12 | /*二叉树中任意两节点的最长路径*/ 13 | 14 | //定义一个初始的最大值 15 | private int max = 0; 16 | 17 | public int diameterOfBinaryTree(TreeNode root) { 18 | depth(root); 19 | return max; 20 | } 21 | 22 | private int depth(TreeNode root) { 23 | if (root == null) return 0; 24 | int leftDepth = depth(root.left); 25 | int rightDepth = depth(root.right); 26 | //左树的长度和右树的长度相加再和最大值比较 27 | max = Math.max(max,leftDepth+rightDepth); 28 | //树的层数 29 | return Math.max(leftDepth, rightDepth) + 1; 30 | } 31 | } 32 | -------------------------------------------------------------------------------- /src/algorithm/_557_Reverse_Words_in_a_String_III/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._557_Reverse_Words_in_a_String_III; 2 | 3 | public class Solution { 4 | 5 | /*给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。*/ 6 | 7 | public String reverseWords(String s) { 8 | char[] str = s.toCharArray(); 9 | int start = 0; 10 | for (int i = 0; i < str.length; i++) { 11 | if (s.charAt(i) == ' ') { 12 | reverseWord(str, start, i - 1); 13 | start = i + 1; 14 | } 15 | } 16 | reverseWord(str, start, str.length - 1); 17 | return new String(str); 18 | } 19 | 20 | private void reverseWord(char[] str, int start, int end) { 21 | while (start < end) { 22 | char temp = str[start]; 23 | str[start] = str[end]; 24 | str[end] = temp; 25 | start++; 26 | end--; 27 | } 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_560_Subarray_Sum_Equals_K/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._560_Subarray_Sum_Equals_K; 2 | 3 | import java.util.HashMap; 4 | 5 | public class Solution { 6 | 7 | /*要求连续子数组的和为k，求子数组的个数*/ 8 | 9 | 10 | 11 | /*采用的方式是用hashmap保存的key是前面数字相加的和，value是该和出现的次数。将该问题转化为前面数字的和减去k后 12 | 的数字和的个数*/ 13 | public int subarraySum(int[] nums, int k) { 14 | int sum = 0, result = 0; 15 | HashMap map = new HashMap<>(); 16 | //当和为0时，次数为1 17 | map.put(0, 1); 18 | for (int i = 0; i < nums.length; i++) { 19 | sum += nums[i]; 20 | //如果sum-k这个key存在，该和出现的次数就表示子数组和为k的个数 21 | if (map.containsKey(sum - k)) { 22 | result += map.get(sum - k); 23 | } 24 | //将前面数字的和作为key放入，map有已经有这个key的话value就加1 25 | map.put(sum, map.getOrDefault(sum, 0) + 1); 26 | } 27 | return result; 28 | } 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_565_Array_Nesting/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._565_Array_Nesting; 2 | 3 | public class Solution { 4 | 5 | /*索引从0开始长度为N的数组A，包含0到N - 1的所有整数。找到并返回最大的集合S， 6 | S[i] = {A[i], A[A[i]], A[A[A[i]]], ... }且遵守以下的规则。 7 | 8 | 假设选择索引为i的元素A[i]为S的第一个元素，S的下一个元素应该是A[A[i]]，之后是 9 | A[A[A[i]]]... 以此类推，不断添加直到S出现重复的元素。*/ 10 | 11 | public int arrayNesting(int[] nums) { 12 | int max = 0; 13 | for (int i = 0; i < nums.length; i++) { 14 | int count = 0; 15 | //从该数字开始访问 16 | for (int j = i; nums[j] != -1; ) { 17 | count++; 18 | int t = nums[j]; 19 | //标记已经使用过 20 | nums[j] = -1; 21 | j = t; 22 | } 23 | max = Math.max(max, count); 24 | } 25 | return max; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_566_Reshape_the_Matrix/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._566_Reshape_the_Matrix; 2 | 3 | public class Solution { 4 | 5 | /*给出一个由二维数组表示的矩阵，以及两个正整数r和c，分别表示想要的重构的矩阵的行数和列数。 6 | 重构后的矩阵需要将原始矩阵的所有元素以相同的行遍历顺序填充。 7 | 如果具有给定参数的reshape操作是可行且合理的，则输出新的重塑矩阵；否则，输出原始矩阵。*/ 8 | 9 | public int[][] matrixReshape(int[][] nums, int r, int c) { 10 | if (nums == null || nums.length == 0) { 11 | return null; 12 | } 13 | int rows = nums.length; 14 | int cols = nums[0].length; 15 | if (rows * cols != r * c) { 16 | return nums; 17 | } 18 | int[][] result = new int[r][c]; 19 | int index = 0; 20 | for (int i = 0; i < r; i++) { 21 | for (int j = 0; j < c; j++) { 22 | result[i][j] = nums[index / cols][index % cols]; 23 | index++; 24 | } 25 | } 26 | return result; 27 | } 28 | } 29 | -------------------------------------------------------------------------------- /src/algorithm/_583_Delete_Operation_for_Two_Strings/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._583_Delete_Operation_for_Two_Strings; 2 | 3 | public class Solution { 4 | 5 | /*给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数， 6 | 每步可以删除任意一个字符串中的一个字符。*/ 7 | 8 | //转化为计算两个字符串的公共字符串 9 | public int minDistance(String word1, String word2) { 10 | int m = word1.length(), n = word2.length(); 11 | int[][] dp = new int[m + 1][n + 1]; 12 | for (int i = 1; i <= m; i++) { 13 | for (int j = 1; j <= n; j++) { 14 | if (word1.charAt(i - 1) == word2.charAt(j - 1)) { 15 | dp[i][j] = dp[i - 1][j - 1] + 1; 16 | } else { 17 | dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); 18 | } 19 | } 20 | } 21 | //两个字符串相加的长度减去公共字符串 22 | return m + n - 2 * dp[m][n]; 23 | } 24 | } 25 | -------------------------------------------------------------------------------- /src/algorithm/_605_Can_Place_Flowers/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._605_Can_Place_Flowers; 2 | 3 | public class Solution { 4 | 5 | /*给定一个数组，数组里面1表示种了花朵，0表示没有种。看能否种下n朵花，要求种的花不能相邻*/ 6 | public boolean canPlaceFlowers(int[] flowerbed, int n) { 7 | int len = flowerbed.length; 8 | int count = 0; 9 | for (int i = 0; i < len && count < n; i++) { 10 | //判断是否种了花 11 | if (flowerbed[i] == 1) { 12 | continue; 13 | } 14 | //左边是否种了 15 | int pre = i == 0 ? 0 : flowerbed[i - 1]; 16 | //右边是否种了 17 | int next = i == len - 1 ? 0 : flowerbed[i + 1]; 18 | //如果左边和右边都没有花就种下 19 | if (pre == 0 && next == 0) { 20 | count++; 21 | flowerbed[i] = 1; 22 | } 23 | } 24 | //数量相等表示都能种下 25 | return count == n; 26 | } 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_617_Merge_Two_Binary_Trees/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._617_Merge_Two_Binary_Trees; 2 | 3 | public class Solution { 4 | 5 | public class TreeNode { 6 | int val; 7 | TreeNode left; 8 | TreeNode right; 9 | TreeNode(int x) { val = x; } 10 | } 11 | 12 | /*给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。 13 | 你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点 14 | 合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。*/ 15 | 16 | public TreeNode mergeTrees(TreeNode t1, TreeNode t2) { 17 | if( t1 == null && t2 == null){ 18 | return null; 19 | }else if( t1 == null){ 20 | return t2; 21 | }else if( t2 == null){ 22 | return t1; 23 | } 24 | TreeNode root = new TreeNode(t1.val + t2.val); 25 | //递归左子树 26 | root.left = mergeTrees(t1.left,t2.left); 27 | //递归右子树 28 | root.right = mergeTrees(t1.right,t2.right); 29 | return root; 30 | } 31 | } 32 | -------------------------------------------------------------------------------- /src/algorithm/_621_Task_Scheduler/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._621_Task_Scheduler; 2 | 3 | import java.util.Arrays; 4 | 5 | public class Solution { 6 | 7 | /*给定一个用字符数组表示的 CPU 需要执行的任务列表。其中包含使用大写的 A - Z 字母表示的26 种 8 | 不同种类的任务。任务可以以任意顺序执行，并且每个任务都可以在 1 个单位时间内执行完。CPU 在任何 9 | 一个单位时间内都可以执行一个任务，或者在待命状态。 10 | 然而，两个相同种类的任务之间必须有长度为 n 的冷却时间，因此至少有连续 n 个单位时间内 CPU 在执 11 | 行不同的任务，或者在待命状态。 12 | 你需要计算完成所有任务所需要的最短时间。*/ 13 | 14 | 15 | public int leastInterval(char[] tasks, int n) { 16 | int[] chars = new int[26]; 17 | for (char c : tasks) { 18 | chars[c - 'A']++; 19 | } 20 | //按频率排序，可以找到使用频率最高的值 21 | Arrays.sort(chars); 22 | int i = 25; 23 | //确定使用频率最高的值相同的字符有多少个 24 | while (i >= 0 && chars[i] == chars[25]) i--; 25 | //考虑间隔长度n可能为零，所以最短时间可能是数组长度 26 | return Math.max(tasks.length, (chars[25] - 1) * (n + 1) + 25 - i); 27 | } 28 | 29 | } 30 | -------------------------------------------------------------------------------- /src/algorithm/_645_Set_Mismatch/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._645_Set_Mismatch; 2 | 3 | public class Solution { 4 | 5 | /*集合 S 包含从1到 n 的整数。不幸的是，因为数据错误，导致集合里面某一个元素复制了成了集合 6 | 里面的另外一个元素的值，导致集合丢失了一个整数并且有一个元素重复。 7 | 给定一个数组 nums 代表了集合 S 发生错误后的结果。你的任务是首先寻找到重复出现的整数，再找到丢 8 | 失的整数，将它们以数组的形式返回。*/ 9 | 10 | //将1-n的整数摆放到正确的位置上 11 | public int[] findErrorNums(int[] nums) { 12 | for (int i = 0; i < nums.length; i++) { 13 | //如果整数位置不正确并且不和交换后的数字相等，就交换数字 14 | while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) { 15 | swap(nums, i, nums[i] - 1); 16 | } 17 | } 18 | for (int i = 0; i < nums.length; i++) { 19 | if (nums[i] != i + 1) { 20 | return new int[]{nums[i], i + 1}; 21 | } 22 | } 23 | return null; 24 | } 25 | 26 | private void swap(int[] nums, int i, int j) { 27 | int temp = nums[i]; 28 | nums[i] = nums[j]; 29 | nums[j] = temp; 30 | } 31 | } 32 | -------------------------------------------------------------------------------- /src/algorithm/_646_Maximum_Length_of_Pair_Chain/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._646_Maximum_Length_of_Pair_Chain; 2 | 3 | import java.util.Arrays; 4 | 5 | public class Solution { 6 | 7 | /*给出 n 个数对。 在每一个数对中，第一个数字总是比第二个数字小。 8 | 现在，我们定义一种跟随关系，当且仅当 b < c 时，数对(c, d) 才可以跟在 (a, b) 后面。 9 | 我们用这种形式来构造一个数对链。 10 | 给定一个对数集合，找出能够形成的最长数对链的长度。你不需要用到所有的数对，你可以以任何顺序 11 | 选择其中的一些数对来构造。*/ 12 | 13 | public int findLongestChain(int[][] pairs) { 14 | if (pairs == null || pairs.length == 0) { 15 | return 0; 16 | } 17 | Arrays.sort(pairs, (o1, o2) -> o1[0] - o2[0]); 18 | int length = pairs.length; 19 | int[] dp = new int[length]; 20 | Arrays.fill(dp, 1); 21 | for (int i = 0; i < length; i++) { 22 | for (int j = 0; j < i; j++) { 23 | if (pairs[j][1] < pairs[i][0]) { 24 | dp[i] = Math.max(dp[i], dp[j] + 1); 25 | } 26 | } 27 | } 28 | return Arrays.stream(dp).max().orElse(0); 29 | } 30 | } 31 | -------------------------------------------------------------------------------- /src/algorithm/_647_Palindromic_Substrings/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._647_Palindromic_Substrings; 2 | 3 | public class Solution { 4 | /*给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。 5 | 具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被计为是不同的子串。*/ 6 | 7 | //回文字符串数量 8 | int count = 0; 9 | 10 | public int countSubstrings(String s) { 11 | for (int i = 0; i < s.length(); i++) { 12 | //字符串长度为奇数，当前结点向外延长 13 | extendPalindrome(s, i, i);//odd 14 | //字符串长度为偶数，当前结点加上后一个结点向外延长 15 | extendPalindrome(s, i, i + 1);//even 16 | } 17 | return count; 18 | } 19 | 20 | //向外延长，继续找出回文字符串 21 | private void extendPalindrome(String s, int left, int right) { 22 | while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) { 23 | count++; 24 | left--; 25 | right++; 26 | } 27 | } 28 | } 29 | -------------------------------------------------------------------------------- /src/algorithm/_669_Trim_a_Binary_Search_Tree/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._669_Trim_a_Binary_Search_Tree; 2 | 3 | public class Solution { 4 | 5 | /*给定一个二叉搜索树，同时给定最小边界L 和最大边界 R。通过修剪二叉搜索树，使得所有节点 6 | 的值在[L, R]中 (R>=L) 。你可能需要改变树的根节点，所以结果应当返回修剪好的二叉搜索树 7 | 的新的根节点。*/ 8 | 9 | public class TreeNode { 10 | int val; 11 | TreeNode left; 12 | TreeNode right; 13 | 14 | TreeNode(int x) { 15 | val = x; 16 | } 17 | } 18 | 19 | public TreeNode trimBST(TreeNode root, int L, int R) { 20 | if (root == null) { 21 | return null; 22 | } 23 | //根节点比左边界小，修剪的树的根节点在右子树中寻找 24 | if (root.val < L) { 25 | return trimBST(root.right, L, R); 26 | } 27 | //根节点比右边界大，修剪的树的根节点在左子树中寻找 28 | if (root.val > R) { 29 | return trimBST(root.left, L, R); 30 | } 31 | //根节点处于L和R中间 32 | root.left = trimBST(root.left, L, R); 33 | root.right = trimBST(root.right, L, R); 34 | return root; 35 | } 36 | } 37 | -------------------------------------------------------------------------------- /src/algorithm/_680_Valid_Palindrome_II/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._680_Valid_Palindrome_II; 2 | 3 | public class Solution { 4 | 5 | /*给定一个非空字符串 s，最多删除一个字符。判断是否能成为回文字符串。*/ 6 | 7 | 8 | public boolean validPalindrome(String s) { 9 | int index = 0, end = s.length() - 1; 10 | while (index < end) { 11 | //如果不匹配，存在两种情况回文 12 | if (s.charAt(index) != s.charAt(end)) { 13 | return isPalindrome(s, index + 1, end) || isPalindrome(s, index, end - 1); 14 | } 15 | index++; 16 | end--; 17 | } 18 | return true; 19 | } 20 | 21 | //双指针判断回文 22 | private boolean isPalindrome(String s, int i, int j) { 23 | while (i < j) { 24 | if (s.charAt(i++) != s.charAt(j--)) { 25 | return false; 26 | } 27 | } 28 | return true; 29 | } 30 | } 31 | -------------------------------------------------------------------------------- /src/algorithm/_739_Daily_Temperatures/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._739_Daily_Temperatures; 2 | 3 | import java.util.Stack; 4 | 5 | public class Solution { 6 | 7 | /*给定一个日常温度的数组T,需要返回一个数组，数组里面记录了还需要等待多少天才能有更高的温度*/ 8 | 9 | public int[] dailyTemperatures(int[] T) { 10 | //用栈来保存每一天 11 | Stack stack = new Stack(); 12 | int[] result = new int[T.length]; 13 | for (int i = 0; i < T.length; i++) { 14 | //栈不是空的并且当前天的温度大于栈顶该天的温度 15 | while (!stack.empty() && T[i] > T[stack.peek()]) { 16 | //栈顶的天数 17 | int index = stack.pop(); 18 | //更新result数组的栈顶天数需要等待的天数 19 | result[index] = i - index; 20 | } 21 | //将天数入栈 22 | stack.push(i); 23 | } 24 | return result; 25 | } 26 | } 27 | -------------------------------------------------------------------------------- /src/algorithm/_746_Min_Cost_Climbing_Stairs/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._746_Min_Cost_Climbing_Stairs; 2 | 3 | public class Solution { 4 | 5 | /*数组的每个索引做为一个阶梯，第 i个阶梯对应着一个非负数的体力花费值 cost[i](索引从0开始)。 6 | 每当你爬上一个阶梯你都要花费对应的体力花费值，然后你可以选择继续爬一个阶梯或者爬两个阶梯。 7 | 您需要找到达到楼层顶部的最低花费。在开始时，你可以选择从索引为 0 或 1 的元素作为初始阶梯。*/ 8 | 9 | 10 | public int minCostClimbingStairs(int[] cost) { 11 | if (cost == null || cost.length == 0) { 12 | return 0; 13 | } 14 | for (int i = 2; i < cost.length; i++) { 15 | cost[i] += Math.min(cost[i - 1], cost[i - 2]); 16 | } 17 | return Math.min(cost[cost.length - 1], cost[cost.length - 2]); 18 | } 19 | } 20 | -------------------------------------------------------------------------------- /src/algorithm/_769_Max_Chunks_To_Make_Sorted/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._769_Max_Chunks_To_Make_Sorted; 2 | 3 | public class Solution { 4 | 5 | /*数组arr是[0, 1, ..., arr.length - 1]的一种排列，我们将这个数组分割成几个“块”，并将 6 | 这些块分别进行排序。之后再连接起来，使得连接的结果和按升序排序后的原数组相同。 7 | 我们最多能将数组分成多少块？*/ 8 | 9 | 10 | public int maxChunksToSorted(int[] arr) { 11 | if (arr == null || arr.length == 0) { 12 | return 0; 13 | } 14 | int right = arr[0]; 15 | int count = 0; 16 | for (int i = 0; i < arr.length; i++) { 17 | right = Math.max(right, arr[i]); 18 | //找到一块逆序，可以来一次排序 19 | if (right == i) { 20 | count++; 21 | } 22 | } 23 | return count; 24 | } 25 | 26 | 27 | } 28 | -------------------------------------------------------------------------------- /src/algorithm/_771_Jewels_and_Stones/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._771_Jewels_and_Stones; 2 | 3 | import java.util.HashSet; 4 | import java.util.Set; 5 | 6 | public class Solution { 7 | 8 | 9 | /* 给定字符串J 代表石头中宝石的类型，和字符串 S代表你拥有的石头。 10 | S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。*/ 11 | 12 | //借助HashSet,时间复杂度为O(M+N) 13 | public int numJewelsInStones(String J, String S) { 14 | int res = 0; 15 | Set set = new HashSet(); 16 | for (char j : J.toCharArray()) { 17 | if (!set.contains(j)) { 18 | set.add(j); 19 | } 20 | } 21 | for (char s : S.toCharArray()) { 22 | if (set.contains(s)) { 23 | res++; 24 | } 25 | } 26 | return res; 27 | } 28 | } 29 | -------------------------------------------------------------------------------- /src/algorithm/_876_Middle_of_the_Linked_List/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._876_Middle_of_the_Linked_List; 2 | 3 | public class Solution { 4 | 5 | 6 | public class ListNode { 7 | int val; 8 | ListNode next; 9 | 10 | ListNode(int x) { 11 | val = x; 12 | } 13 | } 14 | 15 | 16 | 17 | /*给定一个带有头结点 head 的非空单链表，返回链表的中间结点。 18 | 如果有两个中间结点，则返回第二个中间结点。*/ 19 | 20 | public ListNode middleNode(ListNode head) { 21 | ListNode slow = head, fast = head; 22 | while (fast != null && fast.next != null) { 23 | slow = slow.next; 24 | fast = fast.next.next; 25 | } 26 | return slow; 27 | } 28 | } 29 | -------------------------------------------------------------------------------- /src/algorithm/_946_Validate_Stack_Sequences/Solution.java: -------------------------------------------------------------------------------- 1 | package algorithm._946_Validate_Stack_Sequences; 2 | 3 | import java.util.Stack; 4 | 5 | public class Solution { 6 | 7 | /*给定 pushed 和 popped 两个序列，只有当它们可能是在最初空栈上进行的 8 | 推入 push 和弹出 pop 操作序列的结果时，返回 true；否则，返回 false 。*/ 9 | 10 | 11 | //构造辅助栈 12 | public boolean validateStackSequences(int[] pushed, int[] popped) { 13 | if (pushed == null || popped == null) { 14 | return false; 15 | } 16 | Stack stack = new Stack<>(); 17 | //出栈索引位 18 | int popIndex = 0; 19 | for (int i = 0; i < pushed.length; i++) { 20 | stack.push(pushed[i]); 21 | //如果栈不为空且栈顶元素等于出栈序列元素，就执行出栈操作并移动出栈索引 22 | while (!stack.empty() && stack.peek() == popped[popIndex]) { 23 | stack.pop(); 24 | popIndex++; 25 | } 26 | } 27 | //栈为空表示元素全部弹出了，返回true，否则返回false 28 | return stack.empty(); 29 | } 30 | } 31 | --------------------------------------------------------------------------------