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213. House Robber II ​

题目 ​

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 a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

 

Example 1:

Input: nums = [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: nums = [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.

Example 3:

Input: nums = [1,2,3]
Output: 3

 

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 1000
Related Topics
  • 数组
  • 动态规划

  • 👍 1705
  • 👎 0

    • 思路 ​

    将一个环打破，变成两个线性的进行结果对比

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def rob(self, nums: List[int]) -> int:

        n = len(nums)
        if n == 0:
            return 0
        elif n == 1:
            return nums[0]

        def find_rob_linear(nums: List[int]) -> int:
            n = len(nums)
            if n == 0:
                return 0
            elif n == 1:
                return nums[0]


            pre2 = nums[0]

            pre1 = max(pre2, nums[1])

            cur = pre1

            for i in range(2, n):
                cur = max(pre1, nums[i] + pre2 )
                pre2 = pre1
                pre1 = cur

            return cur

        return max(find_rob_linear(nums[:-1]), find_rob_linear(nums[1:]))




# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(1)