Sort Array By Parity python

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.

Return any array that satisfies this condition.

 

Example 1:

Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

• 1 <= nums.length <= 5000
• 0 <= nums[i] <= 5000

 

class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:
        """
        1 <= nums.length <= 5000
        0 <= nums[i] <= 5000
        
        Input: nums = [3,1,2,4]
        Output: [2,4,3,1]
        
        짝수 왼쪽, 홀수 오른쪽
        return => list
        """
        if len(nums) == 1: return nums
        
        x = 0
        y = len(nums) - 1
        result = [0 for _ in range(len(nums))]
        
        for i in range(len(nums)):
            if nums[i] % 2 == 0:
                result[x] = nums[i]
                x += 1
            else:
                result[y] = nums[i]
                y -= 1
                
        return result

성능이 딱 중간입니다.

아래 코드는 거의 정답에 가까운 코드입니다.

class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:
        even = []
        odd = []
        for n in nums:
            if n % 2 == 0:
                even.append(n)
            else:
                odd.append(n)
        return even + odd

짝수, 홀수 리스트를 만들어 합치는 방법을 사용하였습니다.