Day 47 | 198. House Robber | 213. House Robber II | 337. House Robber III

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

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198. House Robber

Question Link

class Solution {
    public int rob(int[] nums) {
        if(nums.length==1)
            return nums[0];
        int[] dp = new int[nums.length];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);
        for(int i = 2; i < nums.length; i++){
            dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i]);
        }
        return dp[nums.length-1];
    }
}

• dp[i]: The maximum amount of money that could be robbed from houses whose index is within the i (include i)
• Recursion Formula
  • If rob the house with index i : dp[i - 2] + nums[i]
  • If don’t rob the house with index i : dp[i - 1]
• Because the recursion formula is based on dp[0] and dp[1]. We could initialize the dp array as follows.
  • According to the definition of dp[i]. The dp[0] must be nums[0].
  • dp[1] is the maximum value of nums[0] and nums[1]
• Traverse order must be from front to back, cause dp[i] is derived from dp[i - 2] and dp[i - 1]

213. House Robber II

Question Link

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 1)
            return nums[0];
        if(nums.length == 2)
            return Math.max(nums[0], nums[1]);
        
        return Math.max(robAction(nums, 0, nums.length - 2), robAction(nums, 1, nums.length - 1));
    }

     public int robAction(int[] nums, int start, int end) {
        int[] dp = new int[nums.length];
        dp[start] = nums[start];
        dp[start+1] = Math.max(nums[start], nums[start+1]);
        for(int i = start + 2; i <= end; i++){
            dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i]);
        }
        return dp[end];
    }
}

• We need to consider two situations:
  • Include the first element, not include the tail element
  • Include the tail element, not include the first element

337. House Robber III

Question Link

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        int[] res = robAction(root);
        return Math.max(res[0], res[1]);
    }

    int[] robAction(TreeNode root) {
        int[] res = new int[2];
        if(root == null)
            return res;
        
        int[] left = robAction(root.left);
        int[] right = robAction(root.right);

        res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        res[1] = root.val + left[0] + right[0];
        return res;
    }
}

• The length of dp[i] array is 2.
  • dp[0] records the maximum amount of money that does not include this node.
  • dp[1] records the maximum amount of money that includes this node.
• Termination condition: null node
• The traversal order must be post order traversal, cause we need the return value of the recursive function to do the next calculation.
• If does not rob current node, then rob left and right child nodes
  • dp[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1])
• If rob current node, then dose not rob left and right child nodes
  • dp[1] = root.val + left[0] + right[0]