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.

其實就是找數組中不相鄰元素的最大和

一道dp的題

用dp的思想來考慮， 對於第N步我們所能有的選擇是偷和不偷：

如果不偷，那我們就不用關心在N-1有沒有偷，所以在N這一步我們能拿到的最大值就是N-1的最大值，即dpnottake[n]=dpmax[n-1]；

如果偷，那麼得到的是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;
    }
};