House Robber II (houses are arranged in a circle)

题目描述：

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.

分析：

由于房子的位置是环状的，所以第一个和最后一个之间有依赖，如果可以打破这个环，就可以用House Robber的方法解了。

对于任意连续的两个房子i，i+1，至少有一个不能偷，所以我们可以将整个事件分为两个独立互补的两个子集：i一定不被偷，i+1一定不被偷。

若i一定不被偷，则可以删掉第i个节点，环断开，可以利用House Robber解，同理解i+1一定不被偷，结果取两者最大值

代码：复杂度：时间：O(n) 空间：O(1)

int rob(vector<int>& nums) {    int n = nums.size();        if( n == 0 )        return 0;    if( n == 1 )        return nums[0];    if( n == 2 )        return max( nums[0], nums[1] );        return max( helper( nums, 0, n-1 ), helper( nums, 1, n ) );}// [left, right)int helper( vector<int>& nums, int left, int right ){    vector<vector<int>> rob( 2, vector<int>(2,0) );        int cur = 0;        int maxrob = 0;        for( int i = left; i < right; ++i )    {        int prv = cur;        cur = cur ^ 1;        rob[cur][0] = max( rob[prv][0], rob[prv][1] );        rob[cur][1] = nums[i] + rob[prv][0];        maxrob = max( rob[cur][0], rob[cur][1] );    }    return maxrob;}