[leetcode]198. House Robber

最新推荐文章于 2025-08-16 00:47:01 发布

你看见我的代码了么

最新推荐文章于 2025-08-16 00:47:01 发布

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分类专栏： LeetCode 文章标签： leetcode DP

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本文介绍了一道经典的动态规划问题——LeetCode 198. 打家劫舍。该问题要求在一个街区中抢劫房屋，但不能连续抢劫相邻的两所房子。文章详细解析了动态规划的思路，并给出了C++实现代码。

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[leetcode]198. House Robber

Analysis

周五啦~—— [高考结束了耶~]

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.

显然是一道动态规划问题，对于第i个房子，如果抢劫那么就不能抢劫第i-1个房子，所以动态转移方程为：dp[i] = max(dp[i-1], dp[i-2]+nums[i-1])。

Implement

class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(len == 0)
            return 0;
        int* dp = new int[len+1];
        dp[0] = 0;
        dp[1] = nums[0];
        for(int i=2; i<=len; i++){
            dp[i] = max(dp[i-1], dp[i-2]+nums[i-1]);
        }
        return dp[len];
    }
};