原題連結

實現原理解析

該題在House Robber的基礎上讓首位連結形成環，那麼即表示第一個和最後一個不能同時被搶，則問題分解為House Robber（nums[0:len(nums)-1]）和House Robber（nums[1:len(nums)]），兩者中比較大的那個即為結果

python程式碼實現

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.dfs(nums[1:]), self.dfs(nums[:-1]))

    def dfs(self, nums):
        if len(nums) == 2:
            return max(nums[0], nums[1
 
])
        res = [0 for each in nums]
        res[0] = nums[0]
        res[1] = max(nums[0], nums[1])
        for index in xrange(2,len(nums)):
            res[index] = max(res[index -2] + nums[index], res[index - 1])
        return res[len(nums) - 1]