Title

the problem link

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.

Solution

题目大意:抢劫银行，不能抢连续的，所以需要跳着来，这个就可以DP动态规划实现，关键代码：

        for(int i=2;i<nums.size();i++){
            nums[i]=max(nums[i]+nums[i-2],nums[i-1]);
        }

以下是完整代码：

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