198. House Robber (dp)

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本文探讨了House Robber问题，即作为专业窃贼如何在不触动相邻房屋警报的情况下，从一系列房屋中窃取最大金额。文章提供了一种动态规划算法解决方案，并详细解释了其实现细节。

198. House Robber

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Total Accepted: 68856 Total Submissions: 199648 Difficulty: Easy

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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

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class Solution {
public:
    int rob(vector<int>& nums) {
       memset(dp,0,sizeof(dp));
       int len=nums.size();
       for(int i=0;i<len;i++)
       {
           dp[i][0]=0;
           dp[i][1]=nums[i];
           if(i>=1) 
           {
               dp[i][0]+=max(dp[i-1][0],dp[i-1][1]);
               dp[i][1]+=dp[i-1][0];
           }
       }
       int ans=0;
       for(int i=0;i<len;i++)
           ans=max(max(dp[i][0],dp[i][1]),ans);
           return ans;
    }
private:
    int dp[1005][2];
};