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.

//和上一題類似，只是多了一個“環”的條件，就是搶了第一家，就不能搶最後一家
//分別計算搶第二家到最後一家與搶第一家到倒數第二家的最大值，取兩個值中更大的那個就是結果。
class Solution {
public:
	int rob(vector<int>& nums) {
		int n = nums.size();
		if (n == 0) return 0;
		if (n == 1) return nums[0];

		int* dp = new int[n + 1];
		//搶第一家到倒數第二家得到的金錢最大值
		dp[0] = nums[0];  
		for (int i = 1; i < n-1 ; ++i){
			dp[i] = max(dp[i - 1],  (i==1? 0: dp[i-2])+nums[i] );
		}
		int res1=dp[n-2];

		//搶第二家到最後一家得到的金錢最大值
		dp[1] = nums[1];
		for (int i = 2; i < n ; ++i){
			dp[i] = max(dp[i - 1], (i == 2 ? 0 : dp[i - 2]) + nums[i]);
		}
		int res2 = dp[n-1 ];

		return max(res1, res2);//返回二者的較大值
	}
};