LeetCode in Python 213. House Robber II

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

与第一题的不同之处在于这个数组是环形的，第一个和最后一个不可以同时偷。其实就是rob了两次第一题，rob(start, end)表示从start到end能偷到的最大值，注意并不一定包括end，但一定包括start；可见house robber i不用把整个dp数组初始，需要的只有两个变量oneBefore和twoBefore，分别代表end=i-1和end=i-2时的最大值，遍历时更新即可。

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0
        if len(nums) == 1: return nums[0]
        
        def robFrom(begin, end):
            # 返回从begin出发到end结束可偷到的最大值 不一定包括end
            oneBefore = twoBefore = res = 0
            for i in range(begin, end):
                temp = max(oneBefore, nums[i] + twoBefore)
                twoBefore = oneBefore
                oneBefore = temp 

            return oneBefore

        return max(robFrom(0, len(nums)-1), robFrom(1, len(nums)))