代码随想录算法训练营第三九天| 198.打家劫舍 213.打家劫舍II 337.打家劫舍 III

今日任务

198.打家劫舍
213.打家劫舍II
337.打家劫舍 III

198.打家劫舍

题目链接： . - 力扣（LeetCode）

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

213.打家劫舍II

题目链接： . - 力扣（LeetCode）

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

337.打家劫舍 III

题目链接： . - 力扣（LeetCode）

/*** 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[] result = robTree(root);return Math.max(result[0], result[1]);}// 0-不偷 1-偷int[] robTree(TreeNode root) {if (root == null) return new int[]{0, 0};int[] left = robTree(root.left);int[] right = robTree(root.right);//偷当前节点，那就不能偷左右节点int val1 = root.val + left[0] + right[0];//不偷当前节点，那左右节点可投可不投，取大值int val2 = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);return new int[]{val2, val1};}    
}