213. House Robber II

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

这次的房子是围成一圈的。。。

0-1-2-3-4-......-（n-1），然后n-1接到了0

如果第0个房子被抢了，那么只需要考虑编号0——n-2的房子，否则，需要考虑1——n-1的房子

这样就把问题分解成了两个子问题：

class Solution {public:    int rob(vector<int>& nums) {        int len=nums.size();        if(len==0)return 0;        else if(len==1)return nums[0];        return max(rob(nums,0,nums.size()-2),rob(nums,1,nums.size()-1));         }    //overload:    //use the function in case I    int rob(vector<int>& nums,int lo,int hi){        int rob=0,notrob=0;        for(int i=lo;i<=hi;i++){            int currob=notrob+nums[i];            notrob=max(notrob,rob);            rob=currob;        }        return max(notrob,rob);    }};