LeetCode 198. House Robber

House Robber

Des

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.

分析

开始以为这个题奇数加一次偶数加一次就行了，后来细看才发现没有那么简单。
这是个动态规划题，要保证不连续取的情况下使最终取得的值最大。这个题网上有很多种解法，最简单的一种是维护sum1 sum2 两个指针，一个当遍历到奇数时加入，一个当遍历到偶数时加入，每次最终的结果要相互比较，这样的好处是，之前取到的值肯定是最大值，并且新加入的肯定和前边的不相邻。

Code

class Solution {
public:
    int rob(vector<int>& nums) {
        int sum1=0,sum2=0;
        for(int i=0;i<nums.size();i++){
            if(i%2==0){
                sum1+=nums[i];
                sum1=max(sum1,sum2);
            }else{
                sum2+=nums[i];
                sum2=max(sum2,sum1);
            }
        }
        return max(sum1,sum2);
    }
};