Day 45 | 70. Climbing Stairs | 322. Coin Change | 279. Perfect Squares

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

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70. Climbing Stairs

Question Link

class Solution {
    public int climbStairs(int n) {
        int[] dp = new int[n+1];
        dp[0] = 1;
        int m = 2;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(i >= j) dp[i] += dp[i-j];
            }
        }
        return dp[n];
    }
}

• dp[i]: The number of combinations that make up the total steps n.
• When we solve for the number of ways to fill the backpack, the recursive formula is:
  • dp[j] += dp[j - i]
• dp[0] must be initialized to 1. If dp[0] = 0, all the value of dp will be 0.
• Question: why do you put n outside and m inside?
  • In this question, if n = 3, 1 step + 2 steps and 2 steps + 1 step are different combinations. So what we are looking for is permutation
  • If we traverse m first, 2 steps can not be in front of 1 step.

322. Coin Change

Question Link

class Solution {
    public int coinChange(int[] coins, int amount) {
        int max = Integer.MAX_VALUE;
        int[] dp = new int[amount + 1];
        // Initilze the dp array to the maximum value
        for(int j = 0; j < dp.length; j++){
            dp[j] = max;
        }
        dp[0] = 0;
        for(int i = 0; i < coins.length; i++){
            for(int j = coins[i]; j <= amount; j++){
                if (dp[j - coins[i]] != max)
                    dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1);
            }
        }
        return dp[amount] == max ? -1 : dp[amount];
    }
}

• dp[j]: The number of fewest coins that make up the j.
• The minimum number of coins that make up the total amount of j - coins[i] is dp[j - coins[i]]
• So the recursive formula is: dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1)
• Array initialization
  • dp[0] = 0 Because the number of coins that make up the total amount of 0 must be 0.
  • According to the recursive formula, dp[j] must be initialized to a max value. Otherwise, dp[j] will be overwritten by the initial value in the process of comparing.
• In this question, we are looking for the fewest number of coins, not the number of permutations or combinations. So no matter what we traverse first, both are fine.
  • We adopt positive order traversal when traversing the capacity. Because in this question, we have an infinite number of each kind of coin. So it a full backpack question.
• Make comparison only when the dp[i-coins[i]] is not the initial maximum value.

322. Coin Change

Question Link

class Solution {
    public int numSquares(int n) {
        int max = Integer.MAX_VALUE;
        int[] dp = new int[n + 1];
        for(int j = 0; j < dp.length; j++)
            dp[j] = max;

        dp[0] = 0;
        for(int i = 1; i*i <= n; i++){      // traverse item
            for(int j = i*i; j <= n; j++){  // traverse capacity
                //if(dp[j - i*i] != max)
                dp[j] = Math.min(dp[j], dp[j - i*i] + 1);
            }
        }
        return dp[n];
    }
}

• dp[j]: The least number of perfect squares numbers that sum to j.
• We don’t need if(dp[j - i*i] != max){ } anymore. Because we can use 1 to make up every n, so it is definitely possible to make up n.