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.
抢劫犯的第二题，第一题题解链接 https://cloud.tencent.com/developer/article/1019125
与第一题不同在于这次是一个换，首尾不能同时抢。
那么可以变通一下，将这个问题转化为0-n-2下标和1-n-1下标两个子问题的最大值问题。
这样就可以保证首尾不会同时被抢，相当于做了两次抢劫犯第一题的过程。
代码语言：javascript复制class Solution {
public:
int rob(vector<int>& nums) {
int a1 = 0;
int b1 = 0;
int a2 = 0;
int b2 = 0;
if(nums.size() == 1) return nums[0];
for(int i = 0; i < nums.size(); i )
{
if(i != nums.size() - 1)
{
if(i & 1) a1 = max(a1 nums[i], b1);
else b1 = max(b1 nums[i], a1);
}
if(i != 0)
{
if(i & 1) a2 = max(a2 nums[i], b2);
else b2 = max(b2 nums[i], a2);
}
}
return max(max(a1, b1), max(a2, b2));
}
};
