leetcode213. [DP]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.
和198题有点不同的是，该列表是循环列表，则a[0]和a[i-1]不能同时选择，仅需两次dp即可，一次为a[0]~a[i-2]一次为a[i-1]~a[1]再subrob下取二者最大值即可～
代码如下：

class Solution(object):    def subrob(self,nums):        Num_=len(nums)        if Num_==1:            return nums[0]        Value=[nums[0],max(nums[0],nums[1])]        for i in range(2,Num_):            Value.append(max(Value[i-2]+nums[i],Value[i-1]))        return max(Value)    def rob(self,nums):        if not len(nums):            return 0        if len(nums)==1:            return nums[0]        num1=nums[:len(nums)-1]        num2=nums[1:len(nums)]        num2.reverse()        max_=self.subrob(num1)        return max(max_,self.subrob(num2))