Middle-题目58：213. House Robber II

题目原文：
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.
题目大意：
接Easy-题目26，但是现在房子围成了环形，仍然求最大利益。
题目分析：
分第一家是否偷讨论，如果偷第一家，那么最后一家不能偷，则退化成第一家到倒数第二家的线性问题，如果不偷第一家，那么相当于从第二家到最后一家的线性问题。
源码：（language：c）

int rob(int* nums, int numsSize) {
    if(numsSize==0)
        return 0;
    else if(numsSize==1)
        return nums[0];
    else if(numsSize==2)
        return nums[0]>nums[1]?nums[0]:nums[1];
    else
        return max(rob1(nums,numsSize-1),rob1(nums+1,numsSize-1));
}
int max(int a,int b) {
    return a>b?a:b;
}
int rob1(int* nums, int numsSize) {
    if(numsSize==0)
        return 0;
    if(numsSize==1)
        return nums[0];
    else if(numsSize==2)
        return nums[0]>nums[1]?nums[0]:nums[1];
    else
    {
        int i;
        int dp1,dp2,dp;

        dp1=nums[0];
        dp2=nums[0]>nums[1]?nums[0]:nums[1];
        for(i=2;i<numsSize;i++)
        {
            dp=dp2>dp1+nums[i]?dp2:dp1+nums[i];
            dp1=dp2;
            dp2=dp;
        }
        return dp;
    }
}

成绩：
1ms，众数100%
Cmershen的碎碎念：
一开始想到对环上每个节点遍历一次，后来发现只讨论第一个节点即可。