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
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1
dp[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;
}
}考虑两种情况
只考虑首或者只考虑尾
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;
}
}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
/**
* 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;
}
}?