Leetcode Weekly Contest 296

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原本 contest 296 的時間是 6/5 12:30，但我人在家樂福直接沒辦法寫，拖著拖著就到今天了。

2293. Min Max Game

Difficulty: Easy, Open in Leetcode

You are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm on nums:

  1. Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
  2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i]as min(nums[2 * i], nums[2 * i + 1]).
  3. For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
  4. Replace the array nums with newNums.
  5. Repeat the entire process starting from step 1.

Return the last number that remains in nums after applying the algorithm.

題目解釋

Solution

照著題目打就過了，這題老實說蠻神奇的，至於時間複雜度：我就暴力！

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class Solution:
    def minMaxGame(self, nums: List[int]) -> int:
        while(len(nums) != 1):
            n = len(nums)
            half = n / 2
            if n == 1:
                return

            newNums = [0] * int(half)

            for i, v in enumerate(nums):
                if i % 2 == 0 and i < half:
                    newNums[i] = min(nums[2 * i], nums[2 * i + 1])
                elif i % 2 == 1 and i < half:
                    newNums[i] = max(nums[2 * i], nums[2 * i + 1])
            nums = newNums
                
        return nums[0]