【LeetCode】213. House Robber II 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/house-robber-ii/description/

题目描述：

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.

题目大意

这个是198. House Robber的拓展。本题目里面的房间是一个环的，也就是说第一个房子和最后一个房子是相邻的。在这种情况下，相邻的两个房子不能一起偷，求能偷到的金额的最大值。

解题方法

这个题多了环的条件，在这个约束下就多了个不同时偷第一个和最后一个就可以了。所以，两种偷的情况：第一种不偷最后一个房间，第二种不偷第一个房间，求这两种偷法能获得的最大值。所以只多了一个切片的过程。

状态转移方程仍然是：

dp[0] = num[0] （当i=0时） 
dp[1] = max(num[0], num[1]) （当i=1时） 
dp[i] = max(num[i] + dp[i - 2], dp[i - 1]) （当i !=0 and i != 1时）

第三个式子就是当前的房间偷的情况和不偷情况下的最大值。

时间复杂度是O(N)，空间复杂度是O(N). (Beats 98%.)

代码如下：

class Solution:
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0
        if len(nums) == 1: return nums[0]
        if len(nums) == 2: return max(nums[0], nums[1])
        N = len(nums)
        return max(self.rob_range(nums[0 : N - 1]), self.rob_range(nums[1 : N]))
    
    def rob_range(self, nums):
        if not nums: return 0
        if len(nums) == 1: return nums[0]
        if len(nums) == 2: return max(nums[0], nums[1])
        N = len(nums)
        dp = [0] * N
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        for i in range(2, N):
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
        return dp[-1]

参考资料：

http://www.cnblogs.com/grandyang/p/4518674.html

日期

2018 年 10 月 9 日 ———— 天气骤冷，注意保暖