Day 43 | 1049. Last Stone Weight II | 494. Target Sum | 474. Ones and Zeroes |

Day 1 | 704. Binary Search | 27. Remove Element | 35. Search Insert Position | 34. First and Last Position of Element in Sorted Array
Day 2 | 977. Squares of a Sorted Array | 209. Minimum Size Subarray Sum | 59. Spiral Matrix II
Day 3 | 203. Remove Linked List Elements | 707. Design Linked List | 206. Reverse Linked List
Day 4 | 24. Swap Nodes in Pairs| 19. Remove Nth Node From End of List| 160.Intersection of Two Lists
Day 6 | 242. Valid Anagram | 349. Intersection of Two Arrays | 202. Happy Numbe | 1. Two Sum
Day 7 | 454. 4Sum II | 383. Ransom Note | 15. 3Sum | 18. 4Sum
Day 8 | 344. Reverse String | 541. Reverse String II | 替换空格 | 151.Reverse Words in a String | 左旋转字符串
Day 9 | 28. Find the Index of the First Occurrence in a String | 459. Repeated Substring Pattern
Day 10 | 232. Implement Queue using Stacks | 225. Implement Stack using Queue
Day 11 | 20. Valid Parentheses | 1047. Remove All Adjacent Duplicates In String | 150. Evaluate RPN
Day 13 | 239. Sliding Window Maximum | 347. Top K Frequent Elements
Day 14 | 144.Binary Tree Preorder Traversal | 94.Binary Tree Inorder Traversal| 145.Binary Tree Postorder Traversal
Day 15 | 102. Binary Tree Level Order Traversal | 226. Invert Binary Tree | 101. Symmetric Tree
Day 16 | 104.MaximumDepth of BinaryTree| 111.MinimumDepth of BinaryTree| 222.CountComplete TreeNodes
Day 17 | 110. Balanced Binary Tree | 257. Binary Tree Paths | 404. Sum of Left Leaves
Day 18 | 513. Find Bottom Left Tree Value | 112. Path Sum | 105&106. Construct Binary Tree
Day 20 | 654. Maximum Binary Tree | 617. Merge Two Binary Trees | 700.Search in a Binary Search Tree
Day 21 | 530. Minimum Absolute Difference in BST | 501. Find Mode in Binary Search Tree | 236. Lowes
Day 22 | 235. Lowest Common Ancestor of a BST | 701. Insert into a BST | 450. Delete Node in a BST
Day 23 | 669. Trim a BST | 108. Convert Sorted Array to BST | 538. Convert BST to Greater Tree
Day 24 | 77. Combinations
Day 25 | 216. Combination Sum III | 17. Letter Combinations of a Phone Number
Day 27 | 39. Combination Sum | 40. Combination Sum II | 131. Palindrome Partitioning
Day 28 | 93. Restore IP Addresses | 78. Subsets | 90. Subsets II
Day 29 | 491. Non-decreasing Subsequences | 46. Permutations | 47. Permutations II
Day 30 | 332. Reconstruct Itinerary | 51. N-Queens | 37. Sudoku Solver
Day 31 | 455. Assign Cookies | 376. Wiggle Subsequence | 53. Maximum Subarray
Day 32 | 122. Best Time to Buy and Sell Stock II | 55. Jump Game | 45. Jump Game II
Day 34 | 1005. Maximize Sum Of Array After K Negations | 134. Gas Station | 135. Candy
Day 35 | 860. Lemonade Change | 406. Queue Reconstruction by Height | 452. Minimum Number of Arrows
Day 36 | 435. Non-overlapping Intervals | 763. Partition Labels | 56. Merge Intervals
Day 37 | 738. Monotone Increasing Digits | 714. Best Time to Buy and Sell Stock | 968. BT Camera
Day 38 | 509. Fibonacci Number | 70. Climbing Stairs | 746. Min Cost Climbing Stairs
Day 39 | 62. Unique Paths | 63. Unique Paths II
Day 41 | 343. Integer Break | 96. Unique Binary Search Trees
Day 42 | 0-1 Backpack Basic Theory（一）| 0-1 Backpack Basic Theory（二）|416. Partition Equal Subset Sum

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1049. Last Stone Weight II

Question Link

class Solution {
    public int lastStoneWeightII(int[] stones) {
        if(stones == null || stones.length == 0)
            return 0;
        int sum = 0;
        for(int stone : stones)
            sum += stone;

        int target = sum / 2;
        int[] dp = new int[target+1];
        for(int i = 0; i < stones.length; i++){
            for(int j = target; j >= stones[i]; j--)
                dp[j] = Math.max(dp[j], dp[j - stones[i]] + stones[i]);
        }
        return sum - 2 * dp[target];
    }
}

• Divide the stones into two piles of equal weight as much as possible. The stone remaining after the collision is the smallest.
• For instance, when input [2,4,1,1], target = (2 + 4 + 1 + 1)/2 = 4, status of 2D array:

	0	1	2	3	4
stones[0]	0	0	2	2	2
stones[1]	0	0	2	2	4
stones[2]	0	1	2	3	4
stones[3]	0	1	2	3	4

494. Target Sum

Question Link

class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int sum = 0;
        for(int num : nums)
            sum += num;
        if((sum + target)%2 != 0) return 0;
        if(target < 0 && -target > sum) return 0;

        int bagSize = (sum+target) / 2;
        int[] dp = new int[bagSize + 1];
        dp[0] = 1;

        for(int i = 0; i < nums.length; i++){
            for(int j = bagSize; j >= nums[i]; j--){
                dp[j] = dp[j] + dp[j - nums[i]];
            }
        }
        return dp[bagSize];
    }
}

• Divide the nums in to an addition set left and a subtraction set right, then we can make the following derivation:
  • left + right = sum
  • left - right = target
  • left - sum + left = target
  • left = (target + sum) / 2
• If sum + target can not be divided exactly by 2, which means that is problem has no solution.
• Status change of 2D array：

	0	1	2	3	4
nums[0]	1	1	0	0	0
nums[1]	1	2	1	0	0
nums[2]	1	3	3	1	0
nums[3]	1	4	6	4	1
nums[4]	1	5	10	10	0

474. Ones and Zeroes

Question Link

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m+1][n+1];
        int oneNum, zeroNum;
        for(String str : strs){
            oneNum = 0;
            zeroNum = 0;
            for(char ch : str.toCharArray()){
                if(ch == '0')
                    zeroNum++;
                else
                    oneNum++;
            }
            // Reverse order traversal
            for(int i = m; i >= zeroNum; i--){
                for(int j = n; j >= oneNum ; j--)
                    dp[i][j] = Math.max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1);
            }
        }
        return dp[m][n];
    }
}

• dp[i][j]：The length of the largest subset of strs with at most i number of 0 and j number of 1
  • Recursion formula: dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1).
  • Compare to the previous recursion formula, zeroNum and oneNum of the string are equivalent to the weight[i], the number of strings themselves corresponds to the value[i].
• The elements of strs are the items in the bag, m and n correspond to the bag capacity.