【LeetCode】213. House Robber ||

213. House Robber II

Description：

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. 

解题思路：

分析：

由于最后一个房子和第一个房子相邻，那么就会出现两种情况：

1）抢劫第一个房子的时候，不能抢劫最后一个房子。这种情况下，就是从输入序列中去除最后一个房子，也就是求输入序列的子序列中抢劫钱数最多，即求序列[6, 6, 0, 3, 1]。

2）抢劫最后一个房子的时候，不能抢劫第一个房子。这种情况下，就是从输入序列去除第一个房子，也就是求输入序列的子序列中抢劫钱数最多，即求序列[6, 0, 3, 1, 9]。

因此：

通过对上面的两种情况的分析，每一种情况都是这道题（【LeetCode】198. House Robber）的输入，我们可以把这道题的结果封装成一个函数，对上面两种情况分别进行输入计算即可。

（1）动态规划解法：

已经AC的代码：

class Solution:

    def __init__(self):
        self.arr = []

    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
            return 0
        if len(nums) == 1:
            return nums[0]
        if len(nums) == 2:
            return max(nums[0], nums[1])
        self.arr = [0 for _ in range(len(nums))]
        return max(self.getRob(nums, self.arr, 0, len(nums)-2),
                   self.getRob(nums, self.arr, 1, len(nums)-1))

    def getRob(self, nums, arr, start, end):
        """
        :param nums: list[int]
        :param arr:  list[int]
        :param start: int
        :param end: int
        :return:
        """
        arr[start] = nums[start]
        arr[start + 1] =max(nums[start], nums[start + 1])
        for i in range(start+2,end + 1):
            arr[i] = max(nums[i] + arr[i - 2], arr[i - 1])
        return  arr[end]

solution = Solution()
# test 1
# nums = []
# test 2
# nums = [2, 3]
# test 3
nums = [2, 3, 2]
print(solution.rob(nums))

​​AC之后，感觉自己写代码不够简洁，参考https://leetcode.com/problems/house-robber-ii/discuss/196728/python-dp-solution，优化代码后为：

已经AC的代码：

class Solution:

    def rob(self, nums):
        """
        :param nums: list[int]
        :return:
        """
        if len(nums) == 0: return  0
        if len(nums) == 1: return  nums[0]
        if len(nums) == 2: return  max(nums)
        def getRob(start, stop, nums):
            n = nums[start: stop + 1]
            dp = [0] * len(n)
            dp[0], dp[1] = n[0], max(n[0], n[1])
            for i in range(2, len(n)):
                dp[i] = max(dp[i - 1], dp[i - 2] + n[i])
            return dp[-1]
        return  max(getRob(0, len(nums) - 2, nums), getRob(1, len(nums) - 1, nums))

solution = Solution()
# test 1
# nums = []
# test 2
# nums = [2, 3]
# test 3
nums = [2, 3, 2]
print(solution.rob(nums))