[Leetcode]House Robber

House Robber
Total Accepted: 45703 Total Submissions: 142461 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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设置两个向量，一个为get，代表取当前元素；一个为pop,代表不取当前元素。

则可以很方便的得到状态转移方程：

get[i] = max(pop[i - 1] + nums[i],get[i - 2] + nums[i]);
pop[i] = max(pop[i - 1],get[i - 1]);

最优解就在这两种情况中存在，再次max()即可。

class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(!len)    return 0;

        vector<int> get(len,0);
        vector<int> pop(len,0);
        get[0] = nums[0];
        int res = nums[0];
        for(int i = 1;i <= len - 1;++ i){
            if(i >= 2){
                get[i] = max(pop[i - 1] + nums[i],get[i - 2] + nums[i]);
            }
            else
                get[i] = pop[i - 1] + nums[i];
            pop[i] = max(pop[i - 1],get[i - 1]);
            res = max(res,max(get[i],pop[i]));
        }
        return res;
    }
};