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.

题意：假设所有的房子都相连组成一个圈，仍然不能取相邻的房子，那么最大收益为如何。

思路：原先房子是一列，则动态规划求解出最大值，现在房子是一个圈，则第1个房子和第n个房子不能同时取，那么问题就分为两大类，有1没n的时候最大值，和有n没1的时候的最大值，即求出sum(0, n-2)与sum(1, n-1)并比较大小即可。

class Solution {
public:
	int rob(vector<int>& nums) {
		if (nums.empty()){
			return 0;
		}
		int length = nums.size();
		if (length == 1)
			return nums[0];

		//calaulate 0 to n-2
		vector<int> f(length, 0);
		f[1] = nums[0];
		for (int i = 2; i < length; i++){
			f[i] = max(f[i - 1], f[i - 2] + nums[i - 1]);
		}
		
		if (length == 2)
			return max(nums[0], nums[1]);
		//calculate 1 to n-1
		vector<int> g(length, 0);
		g[1] = nums[1];
		for (int i = 2; i < length; i++){
			g[i] = max(g[i - 1], g[i - 2] + nums[i]);
		}

		return max(f[length - 1], g[length - 1]);
	}
};