Day 52 | 300. Longest Increasing Subsequence | 674. Longest Continuous Increasing Subsequence | 718.

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

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300. Longest Increasing Subsequence

Question Link

class Solution {
    public int lengthOfLIS(int[] nums) {
        int[] dp = new int[nums.length];
        Arrays.fill(dp, 1);
        int res = 1;
        for(int i = 1; i < nums.length; i++){
            for(int j = 0; j < i; j++)
                if(nums[i] > nums[j]) dp[i] = Math.max(dp[i], dp[j] + 1);
            // Choose the longest one
            if(dp[i] > res)
                res = dp[i];
        }
        return res;
    }
}

• dp[i]: The length of the longest strictly increasing subsequence ending innums[i] that includes i and before i.
• Recursive Formula：if(nums[i] > nums[j]) dp[i] = Math.max(dp[i], dp[j] + 1);
• For each i, the initial value of dp[i] is at least 1
• Traversal order is from front to back

674. Longest Continuous Increasing Subsequence

Question Link

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int[] dp = new int[nums.length];
        Arrays.fill(dp, 1);
        int res = 1; 
        for(int i = 1; i < nums.length; i++){
            if(nums[i] > nums[i-1])
                dp[i] = dp[i-1] + 1; 
            
            res = Math.max(dp[i], res);
        }
        return res;
    }
}

• dp[i]: The length of the longest continuous increasing subsequence ending in nums[i].
• If nums[i] > nums[i-1], the length of the longest continuous increasing subsequence ending in nums[i] must be equal to the length of the longest continuous increasing subsequence ending in nums[i-1] + 1

718. Maximum Length of Repeated Subarray

Question Link

class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int result = 0;
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];

        for(int i = 1; i < nums1.length + 1; i++){
            for(int j = 1; j < nums2.length + 1; j++){
                if(nums1[i-1] == nums2[j-1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    result = Math.max(result, dp[i][j]);
                }
            }
        }
        return result;
    }
}

• dp[i][j]: The maximum length of repeated subarray of nums1 ends with [i-1] and nums2 ends with [j-1].
• According to to the definition of dp[i][j], it can only be deduced from dp[i-1][j-1]. So when nums1[i - 1] and nums2[j - 1] are equal, the recursion formula is as follows:
  • dp[i][j] = dp[i-1][j-1] + 1
  • According to the recursion formula, we should start from 1 when traversing i and j.
• dp[i][0] and dp[0][j] both are actually meaningless. But for the convenience of recursive formula, we should initialize it to 0.