题目

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.

题目来源：https://leetcode.com/problems/house-robber/

分析

动态规划，一维辅助数组dp。dp[i]表示从第i个元素开始往右，所能抢劫到的最大财富值（抢劫了第i个）。从右往左遍历，dp[i] = nums[i] + max(dp[i+2], dp[i+3]…dp[len-1])。可以设置一个辅助变量，维护自dp[i+2]往右的最大值，省得每次都求了。
还可以优化一下空间，我觉得可以不用一维辅助数组，用三个变量就行了。

代码

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