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.

解题思路：很巧妙地把题目按照第i个位置是否robber把原来的环变成两个子序列，转换成了之前的直线状态下的robber解决方案

class Solution {
public:
    int robber(vector<int>& nums, int l, int r) {
        int pre = 0, cur = 0;
        for (int i = l; i <= r; i++) {
            int temp = max(pre + nums[i], cur);
            pre = cur;
            cur = temp;
        }
        return cur;
    }
    int rob(vector<int>& nums) {
        int n=nums.size();
        if(n<2)return n?nums[0]:0;
        return max(robber(nums,0,n-2),robber(nums,1,n-1));
    }
};