198. House Robber

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 tonightwithout alerting the police.

        这个题用动态规划来解决，定义F(n)为前n个房子所得的最高价值。

        那么其状态转移方程就是：F(n)=max{ F(n-1)，F(n-2)+Prices[n] }。其意思是前n个房子所得的最高价值等于在前n-1个房子所得的最高价值和第n个房子+前n-2个房子所得最高价值中取最大值。有了这个方程就很好编程了。

class Solution {public:    int rob(vector<int> &num) {        int n = num.size();        int p1,p2;        if (n < 2) return n ? num[0] : 0;        else        {            p1 = num[0];            p2 = max(num[0], num[1]);            for(int i = 2; i < n; i ++){                int m = max(p1+num[i], p2);                p1=p2;                p2=m;            }            return p2;        }    }};