[leetcode 198]House Robber

Question:

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:

<span style="font-size:14px;">class Solution {public:    int rob(vector<int>& nums) {        /*        采用动态规划法        fk[i]表示第i个阶段（房子数量为i）时候的最大钱数，可以知道fk[0] = 0; fk[1]=nums[0];        fk[2] = max(nums[0],nums[1]) = max(fk[0]+nums[1],fk[1]);        多列几个阶段的策略即可看出规律！----fk[i] = max((fk[i-2]+nums[i-1]),fk[i-1]);其中nums[i-1]对应第i个房子里的钱。        */        const int size = nums.size();        vector<int> fk(size+1);        if(nums.empty())            return 0;        else{            if(nums.size() == 1)                return nums[0];            else{                fk[0] = 0;                fk[1] = nums[0];                for(int i = 2; i <= size; i++){                    fk[i] = max((fk[i-2]+nums[i-1]),fk[i-1]);                }                return fk[size];            }        }    }};</span>