LeetCode OJ 之 House Robber II （抢劫犯 - 二）

题目：

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.

在上一题的基础上房子围成了一个圆圈。求可以抢劫的最多钱数。

思路：

先参考抢劫犯（1）：http://blog.csdn.net/u012243115/article/details/46340353 。

数组循环意味着首尾相邻，即nums[0]，nums[len-1]相邻，即如果抢劫nums[0]就不能抢劫nums[len-1]，因此可以分两种情况：

1、抢劫nums[0]，而不抢劫nums[len-1]

2、抢劫nums[len-1]，而不抢劫nums[0]。

然后取两个结果的较大值即可。

代码：

class Solution {
public:
int rob(vector<int>& nums)
{
int len = nums.size();
if(len == 0)
return 0;
//注意要单独考虑只有一个数字的情况
if(len == 1)
return nums[0];

int f0 = 0 , f1 = 0 , f2 = 0;
for(int i = 0 ; i < len-1 ; i++)
{
f2 = max(f0 + nums[i] , f1);
f0 = f1;
f1 = f2;
}

int g0 = 0 , g1 = 0 , g2 = 0;
for(int i = 1 ; i < len ; i++)
{
g2 = max(g0 + nums[i] , g1);
g0 = g1;
g1 = g2;
}
return max(f2 , g2);
}
};