题目描述
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.
这道题目的要求可以归纳为数组不相邻元素的最大和。这道题目可以根据动态规划的思想去解答。
对数组num从1…n的最大和f(n)可能有以下两种情况（注意这里为了说明方便，假设数组的起始元素为1）：
f(n) = f(n-2) + num[n]；f(n) = f(n-1)；
f(n)取以上两个值的最大值。当然本题可以使用递归的思路去处理，但是这样会进行多次重复的运算（本题其实类似于斐波那契数列），因此，从前向后的处理更有效率，这里只需要保存f(n-1)和f(n-2)的值即可。
代码如下：
class Solution {public:int rob(vector<int> &num) {int size = num.size();if(size == 0)return 0;if(size == 1)return num[0];if(size == 2)return max(num[0], num[1]);int n1 = num[0];int n2 = num[0] > num[1] ? num[0] : num[1];for(int i = 2; i != num.size(); ++i){int n = n1 + num[i];int n3 = n > n2 ? n : n2;n1 = n2;n2 = n3;}int r = max(n1, n2);return r;}};
，偶尔为街头独特的风景驻足，
