Day 58 | 739. Daily Temperatures | 496. Next Greater Element I

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
Day 43 | 1049. Last Stone Weight II | 494. Target Sum | 474. Ones and Zeroes
Day 44 | Full Backpack Basic Theory | 518. Coin Change II | 377. Combination Sum IV
Day 45 | 70. Climbing Stairs | 322. Coin Change | 279. Perfect Squares
Day 46 | 139. Word Break | Backpack Question Summary
Day 48 | 198. House Robber | 213. House Robber II | 337. House Robber III
Day 49 | 121. Best Time to Buy and Sell Stock I | 122. Best Time to Buy and Sell Stock II
Day 50 | 123. Best Time to Buy and Sell Stock III | 188. Best Time to Buy and Sell Stock IV
Day 51 | 309. Best Time to Buy and Sell Stock with Cooldown | 714. with Transaction Fee
Day 52 | 300. Longest Increasing Subsequence | 674. Longest Continuous Increasing Subsequence | 718. Maximum Length of Repeated Subarray
Day 53 | 1143. Longest Common Subsequence | 1035. Uncrossed Lines | 53. Maximum Subarray
Day 55 | 392. Is Subsequence | 115. Distinct Subsequences
Day 56 | 583. Delete Operation for Two Strings | 72. Edit Distance
Day 57 | 647. Palindromic Substrings | 516. Longest Palindromic Subsequence

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Monotonic Stack

• In a one-dimensional array, find the first element on the right or left of an element that is larger or smaller than itself.
• The time complexity is O(n).
• The monotonic stack only needs to hold the index of the element.
• There are three main criteria for using the monotonic stack.
  • 1、The current element T[i] is smaller than the stack top element T[stack.top()].
  • 2、The current element T[i] is the same as the stack top element T[stack.top()].
  • 3、The current element T[i] is larger than the stack top element T[stack.top()].
• When the monotonic stack is increasing (from the top to the bottom of the stack), this question is to find the one on the right that is bigger than the current element.
• When the monotonic stack is decreasing (from the bottom to the top of the stack), this question is to find the one on the right that is smaller than the current element.

739. Daily Temperatures

Question Link

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int lens = temperatures.length;
        int[] res = new int[lens];
        Deque<Integer> stack = new LinkedList<>();
        stack.push(0);
        for(int i = 1; i < lens; i++){
            if(temperatures[i] <= temperatures[stack.peek()]){
                stack.push(i);
            }else{
                while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]){
                    res[stack.peek()] = i - stack.peek();
                    stack.pop();
                }
                stack.push(i);
            }
        }
        return res;
    }
}

• When the monotonic stack is increasing (from the top to the bottom of the stack), this question is to find the one on the right that is bigger than the current element.

496. Next Greater Element I

Question Link

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Stack<Integer> stack = new Stack<>();
        stack.push(0);

        int[] res = new int[nums1.length];
        Arrays.fill(res, -1);

        HashMap<Integer, Integer> hashMap = new HashMap<>();
        for(int i = 0; i < nums1.length; i++)
            hashMap.put(nums1[i], i);

        for(int i = 1; i < nums2.length; i++){
            if(nums2[i] <= nums2[stack.peek()])
                stack.add(i);
            else {
                while(!stack.isEmpty() && nums2[i] > nums2[stack.peek()]){
                    if(hashMap.containsKey(nums2[stack.peek()])){
                        int index = hashMap.get(nums2[stack.peek()]);
                        res[index] = nums2[i];
                    }
                    stack.pop();
                }
                stack.add(i);
            }
        }
        return res;
    }
}

• This question is similar like the 739. Daily Temperatures. Keep the elements of the stack in increasing order(from the top to the bottom of the stack), then we can get the first element on the right that is larger than itself.
• In the nums2 traversal progress, when the current element is larger than the top element of the stack. We should determine if the top element of the stack has appeared in the nums1. If it appears, record the result.