Day 32 | 122. Best Time to Buy and Sell Stock II | 55. Jump Game | 45. Jump Game II

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

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LeetCode 122. Best Time to Buy and Sell Stock II

Question Link

class Solution {
    public int maxProfit(int[] prices) {
        int result = 0;
        for(int i = 1; i < prices.length; i++){
            int profit = prices[i] - prices[i - 1];
            if(profit > 0)
                result += profit;
        }
        return result;
    }
}

• We can decompose the final profit by day. The max profit is the sum of the profit of days whose profit is positive.

LeetCode 55. Jump Game

Question Link

class Solution {
    public boolean canJump(int[] nums) {
        int coverRange = 0;
        for(int i = 0; i <= coverRange; i++){
            coverRange = Math.max(coverRange, i + nums[i]);
            if(coverRange >= nums.length - 1)
                return true;
        }

        return false;
    }
}

• The initial cover range should be 0
• We only can update the cover range when i in the cover range

LeetCode 45. Jump Game II

Question Link

class Solution {
    public int jump(int[] nums) {
        if(nums == null || nums.length == 0 || nums.length == 1)
            return 0;
        // The number of jumps
        int count = 0;
        // The current cover range
        int curDistance = 0;
        // The max cover range
        int maxDistance = 0;
        for(int i = 0; i < nums.length; i++){
            maxDistance = Math.max(maxDistance, i + nums[i]);
            // Jump one more step can reach the end position.
            if(maxDistance >= nums.length - 1){
                count++;
                break;
            }
            // Update the distance that the next step can reach when reaching the current max cover range
            if(curDistance == i){
                count++;
                curDistance = maxDistance;
            }
        }
        return count;
    }
}

• Jump one more step can reach the end position.
• Update the distance that the next step can reach when reaching the current max cover range