213. House Robber II［Medium］

题目大意

Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street. 

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.

是之前抢劫房子的升级版，现在假设小偷要偷的house连成了一个环，同样不能抢相邻两户，问最多能抢的钱。

解题思路

之前的动态转移方程是 针对直线的 f［i］ ＝ max（f［i－2］＋ v［i］，f［i－1］）；我们可以把环 拆成直线，假设编号 1～5，则1与5相邻，因此有两种直线情况：

1）抢劫1，不能抢劫5，直线1～4

2）不抢劫1，能抢5，直线2～5

用两个数组去分别考虑，然后返回两数组中最大值（这是一遍AC题，开心）

代码如下

class Solution {public:    int rob(vector<int>& nums) {if(nums.empty())return 0;if(nums.size() == 1)return nums[0];vector<int> re1(nums.size(),0);vector<int> re2(nums.size(),0);re1[0] = nums[0];re1[1] = nums[0];re2[0] = 0;re2[1] = nums[1];for(int i = 2; i < nums.size()-1; i++){re1[i] = fmax(nums[i]+re1[i-2], re1[i-1]);}for(int i = 2; i < nums.size(); i++){re2[i] = fmax(nums[i]+re2[i-2], re2[i-1]);}return fmax(re1[nums.size()-2], re2[nums.size()-1]);}};