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.

这题是House Robber的follow up,加入了环的考虑。其实限制就是如果我们取数组的头，则一定没有办法取数组的尾元素，如果取了尾则一定没有办法取数组的头元素。即我们不一定取全部数组的最大rob结果。而是要么取nums[1...n], 要么是nums[0...n-1]。最后取两个结果的最大值，结合house robber的第一题，则可以得出本题的解法，时间复杂度为O(n),空间复杂度为O(1),代码如下：

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        #1.first pop out the leftest element
        #2.next pop out the rightest elemnt
        #3.get the max value of them two.
        if not nums:
            return 0
        if len(nums) < 3:
            return max(nums)
        #first pop out the leftest element
        f1 = nums[1]
        f2 = max(f1,nums[2])
        for n in nums[3:]:
            res = max(f1 + n, f2)
            f1 = f2
            f2 = res
        res1 = f2
        
        f1 = nums[0]
        f2 = max(f1,nums[1])
        for n in nums[2:-1]:
            res = max(f1 + n, f2)
            f1 = f2
            f2 = res
        res2 = f2
        return max(res1, res2)

 

转载于:https://www.cnblogs.com/sherylwang/p/5579987.html