leetcode 198. House Robber

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.

其实就是找数组中不相邻元素的最大和

一道dp的题

用dp的思想来考虑， 对于第N步我们所能有的选择是偷和不偷：

如果不偷，那我们就不用关心在N-1有没有偷，所以在N这一步我们能拿到的最大值就是N-1的最大值，即dpnottake[n]=dpmax[n-1]；

如果偷，那么得到的是N-1不偷的值加上nums[n]的值，即dptake[n]=dpnottake[n-1]+nums[n]。 

第N步的最大值就是dpmax[n] = max(dptake[n], dpnottake[n])

class Solution {
public:
    int rob(vector<int>& nums) {
        int dptake = 0;
        int dpnottake = 0;
        int dpmax = 0;
        for(int i = 0; i < nums.size(); i++){
            dptake = dpnottake + nums[i];
            dpnottake = dpmax;
            dpmax = max(dptake, dpnottake);
        }
        return dpmax;
    }
};