代码随想录二刷day48

乱世在摸鱼2023-10-07 14:45

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文章目录

前言
[一、力扣198. 打家劫舍](#一、力扣198. 打家劫舍)
[二、力扣213. 打家劫舍 II](#二、力扣213. 打家劫舍 II)
[三、力扣337. 打家劫舍 III](#三、力扣337. 打家劫舍 III)

前言

一、力扣198. 打家劫舍

java 复制代码

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

java 复制代码

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

java 复制代码

/**
 * 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;
    }
}