Day 31 | 455. Assign Cookies | 376. Wiggle Subsequence | 53. Maximum Subarray

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

---

LeetCode 455. Assign Cookies

Question Link

class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int count = 0;
        int indexG = g.length-1;
        int indexS = s.length-1;
        while(indexS >= 0 && indexG >=0){
            if(s[indexS] >= g[indexG]){
                count++;
                indexS--;
            }
            indexG--;
        }
        return count;
    }
}

• Satisfy the children that have a bigger greed factor first

LeetCode 376. Wiggle Subsequence

Question Link

class Solution {
    public int wiggleMaxLength(int[] nums) {
        int count = 1;
        int preDiff = 0;
        int curDiff = 0;
        
        for(int i = 1; i < nums.length; i++){
            // Get current difference
            curDiff = nums[i] - nums[i-1];

            // The initial value of preDiff is 0, so we should include this situation
            // If the current difference and the previous difference are one positive and one negative
            if(curDiff < 0 && preDiff >= 0 || curDiff > 0 && preDiff <= 0){
                count++;
                preDiff = curDiff;
            }
        }   
        return count;
    }
}

• If we want a wiggle subsequence, remove the element on the monotonic interval

LeetCode 53. Maximum Subarray

Question Link

class Solution {
    public int maxSubArray(int[] nums) {
        int sum = Integer.MIN_VALUE;
        int count = 0;
        for(int num : nums){
            count += num;
            sum = Math.max(sum, count);
            if(count < 0)
                count = 0;
        }
        return sum;    
    }
}

• Use Math.max(sum, count) to select the max value in the interval. That is, keep finding the terminal position of the maximum subarray.
• When count < 0, reset the start position and count. Because negative numbers definitely lower the sum.