15算法课程 198. House Robber

最新推荐文章于 2025-08-17 21:40:23 发布

QingJiuYou

最新推荐文章于 2025-08-17 21:40:23 发布

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文章标签：算法

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针对一个专业窃贼计划抢劫一排房屋的问题，每个房屋都藏有一定数量的钱财，但相邻的房屋安装有联动报警系统。文章介绍了一种通过动态规划算法解决此问题的方法，旨在找出在不触发报警的情况下能获得的最大金额。

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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.

solution：

由于抢劫的房间不能相邻，所有我们可以看到加入nums[i]与不加入nums[i]的区别。
我们找到递归公式dp[i] = max(dp[i-1],dp[i-2] + nums[i]);

code:

class Solution {
public:
    int rob(vector<int>& nums) {
        vector<int> dp(nums.size(),0);

        if(nums.size() <= 0){
            return 0;
        }
        dp[0] = nums[0];
        for(int i = 1;i < nums.size();++i){
            if( i >= 2){
               dp[i] = max(dp[i-1],dp[i-2]+nums[i]); 
            }else{
               dp[i] = max(dp[i-1],nums[i]);
            }
        }
        return dp[nums.size()-1];
    }
};