不要忘记空数组和数组长度为1的情况单独考虑

和前两个状态有关

代码随想录

class Solution {public int rob(int[] nums) {if(nums == null && nums.length == 0) return 0;if(nums.length == 1) return nums[0];int[] dp = new int[nums.length];//int[] dp = new int[2];dp[0] = nums[0];//和前两个状态有关 所以初始化前两个0和1dp[1] = Math.max(nums[0], nums[1]);//int res;for(int i = 2; i < nums.length; i++) {dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i]);//考虑前一个与考虑前两个//res = Math.max(dp[1], dp[0] + nums[i]);//dp[0] = dp[1];//dp[1] = res;}return dp[nums.length - 1];//return res;}
}

 213.打家劫舍II  

考虑两种情况

只考虑首或者只考虑尾

代码随想录

class Solution {public int rob(int[] nums) {if (nums == null || nums.length == 0)return 0;if (nums.length == 1)return nums[0];int num1 = robNum(nums, 0, nums.length - 1);//只考虑头不考虑尾int num2 = robNum(nums, 1, nums.length);//不考虑头只考虑尾return Math.max(num1, num2);//两者取最大}public int robNum(int[] nums, int start, int end) {int x = 0, y = 0, z = 0;for(int i = start; i < end; i++) {z = Math.max(y, x + nums[i]);x = y;y = z;}return z;}
}

 337.打家劫舍III  

好难

二叉树 后序遍历 递归

递归三部曲：1.参数和返回值 2.终止条件 3.单层递归逻辑

代码随想录

/*** 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]);//取大值}public 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]);//不偷res[1] = root.val + left[0] + right[0];//偷return res;}
}