House Robber

题目

原题

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.

思路

题目大概意思是在一个数组里面, 找出部分/所有不相邻的数, 使得它们的和最大. 显然用DP来用比较方便和直观. dp[i] = max(dp[i-2]+nums[i], dp[i-1]) dp[i]表示以下标i为结尾的所有/部分不相邻的数的和的最大值. 这样就可以通过dp[i-2]+nums[i]和dp[i-1]来递推. 即选择或者不选择nums[i]的问题. (想想为什么? 因为nums[i] >= 0是恒成立的) 剩下的就是边界问题和初始化的事情了.

code

class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(len <= 0) return 0;
        if(len <= 1) return nums[0];

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