題目描述：

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.

題目思路：

1.     先不考慮環的情況，即先處理此題的原始版本

2.     求解無環時，問題的關鍵在於針對某一間房子搶和不搶的選擇。

假如要搶第i個, 就不能搶第i+1個;

假如不搶第i個,那麼可以選擇搶或者不搶第i+1個

很明顯，這是一個動態規劃問題。

3.     可以設定陣列result[],result[i]表示到該房子時最優選擇下的金錢數

4.     加上環的條件後，考慮兩種情況即可。第一是搶了第一家，不搶最後一家，第二是不搶第一家，即分別計算搶了2-n和搶了1-n-1的情況，取最大值就是結果

動態演算法：

       result[i] = max(result[i - 1],  result[i - 2] + nums[i]);

程式碼如下：

class Solution(object): 
	def rob(self, nums):
		 """ :type nums: List[int] :rtype: int """
		 length=len(nums)
		 if length==0:
			 return 0
		 elif length==1:
			 return nums[0]
		 elif length==2:
			 return max(nums[0],nums[1])
		 return max(self.dp(nums[1:]),self.dp(nums[:-1]))
	def dp(self,nums):
		if len(nums)==2:
			return max(nums[0],nums[1])
		result=[0 for each in nums]
		result[0]=nums[0]
		result[1]=max(nums[0],nums[1])
		for i in range(2,len(nums)):
			result[i]=max(result[i-2]+nums[i],result[i-1])
		return result[len(nums)-1] 

結果如下：