LeetCode - 198. House Robber

最新推荐文章于 2025-04-13 23:45:03 发布

热爱满天星

最新推荐文章于 2025-04-13 23:45:03 发布

阅读量255

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签： LeetCode 198 House Robber

本文链接：https://blog.youkuaiyun.com/SMonkeyKing/article/details/82963322

LeetCode 专栏收录该内容

16 篇文章

订阅专栏

本文探讨了一种算法挑战，即如何在不触发相邻房屋安全系统的情况下，从一系列藏有不同金额的房屋中，计算出可抢夺的最大金额。通过动态规划的方法，详细解释了如何实现这一目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Example

Example 1:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.

Example 2:
Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

Code

你是一个专业的强盗，计划在街上抢劫房屋。每个房子都藏着一定数量的钱，阻止你抢劫他们的唯一限制因素是相邻的房屋有连接的安全系统，如果两个相邻的房子在同一个晚上被闯入，它将自动联系警方。给出一个代表每个房子的金额的非负整数列表，确定今晚可以抢劫的最大金额而不警告警察。

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0)
        {
            return 0;
        }
    
        if(nums.length == 1)
        {
            return nums[0];
        }
        // 设置一个dp数组
        int dp[] = new int[nums.length];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0],nums[1]);
        int max = dp[1];
        for(int i = 2;i<=nums.length-1;i++)
        {
            dp[i] = Math.max(dp[i-2]+nums[i],dp[i-1]);
            max = Math.max(max,dp[i]);
        }        
        return max;      
    }
}