House Robber问题及解法

问题描述：

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.

问题分析：

作为一个强盗，去盗取一条街上住户家里的钱，但不能同时盗取相邻的两家的钱。

我么可以这么定义状态：dp[i]----走到第i个住户时，能盗取的钱的最大数目。

所以dp[i] = max(dp[i - 1], dp[i - 2] + nums[i - 1])，最终求得的dp[n]即为答案。

过程详见代码：

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