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前言

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一、力扣198. 打家劫舍

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]);}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-2] + nums[i], dp[i-1]);}return dp[dp.length-1];}
}

二、力扣213. 打家劫舍 II

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]);}int a = fun(nums, 0, nums.length-2);int b = fun(nums, 1, nums.length-1);return Math.max(a,b);}public int fun(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-2]+nums[i], dp[i-1]);}return dp[end];}
}

三、力扣337. 打家劫舍 III

/*** 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 = fun(root);return Math.max(res[0], res[1]);}public int[] fun(TreeNode root){int[] arr = new int[2];if(root == null){return arr;}//0不偷， 1 偷int[] left = fun(root.left);int[] right = fun(root.right);arr[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);arr[1] = root.val + left[0] + right[0];return arr;}
}