198. House Robber

本文详细解析了House Robber问题的解决方法,该问题要求在不连续抢劫相邻房屋的情况下,计算出能获得的最大金额。通过将问题分解为多个子问题,并使用动态规划的方法求解最大值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

198. House Robber

题目:
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.
代码如下:
#define max(a, b) ((a > b) ? a : b)

class Solution {
public:
    int rob(vector<int>& nums) {
        int case1 = 0, case2 = 0;
        for (int i = 0; i < nums.size(); i++) {
          if (i % 2 == 0)
            case1 = max(case1 + nums[i], case2);
          else 
            case2 = max(case1, case2 + nums[i]);
        }

        return max(case1, case2);
    }
};
解题思路:

本题初看复杂,仔细思考实则简单。主要问题是把问题分解成多个子问题,即求出到目前房子为止,之前所有房子所能抢到的最大值,然后继续这个过程。

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值