198. House Robber

本文介绍了一个模拟夜间抢劫房屋的算法问题。该问题设定为一名专业窃贼计划抢劫沿街的房屋,但相邻房屋的安全系统相互连接,若同一晚被连续闯入会报警。通过给定每个房屋藏匿的金钱数额,算法需找出最大抢劫金额的同时不触发报警。

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

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.


class Solution {
public:
    int rob(vector<int>& nums) {
        int x = 0, y = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (i & 1) {
                x = max(x + nums[i], y);
            } else {
                y = max(y + nums[i], x);
            }
        }
        return max(x, y);
    }
};


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