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[5/2] 905. Sort Array By Parity

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작성일 2022.05.01 18:23

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[LeetCode 시즌 3] 2022년 5월 2일 문제입니다.

https://leetcode.com/problems/sort-array-by-parity/

[Easy] 905. Sort Array By Parity

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

#LeetCode, #Facebook, #Microsoft, #Google, #Amazon, #Array, #Two Pointers, #Sorting

링크

https://leetcode.com/problems/sort-array-by-parity/ 2050 회 연결

익명
작성일 2022.05.02 11:49

Runtime: 117 ms, faster than 37.62% of Python3 online submissions for Sort Array By Parity.
Memory Usage: 14.6 MB, less than 63.70% of Python3 online submissions for Sort Array By Parity.

class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:
        p1, p2 = 0, len(nums) - 1
        while p1 < p2:
            if nums[p1] % 2 == 1 and nums[p2] % 2 == 0:
                nums[p1], nums[p2] = nums[p2], nums[p1]    
            while p1 < len(nums) and nums[p1] % 2 == 0:
                p1 += 1
            while p2 >= 0 and nums[p2] % 2 == 1:
                p2 -= 1
        return nums

익명
작성일 2022.05.02 17:29

Runtime: 162 ms
Memory Usage: 46.3 MB

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortArrayByParity = function(nums) {
  parityArray = []
  for (var i = 0; i < nums.length; i++) {
    if (nums[i] % 2 === 0) {
      parityArray.push(nums.splice(i,1))
      i--;
    }
  }
  for (var i = 0; i < nums.length; i++) {
    parityArray.push(nums[i]);
  }
  return parityArray.flat()
};

익명
작성일 2022.05.02 19:17

C++
Runtime: 15 ms, faster than 48.12% of C++ online submissions for Sort Array By Parity.
Memory Usage: 16.3 MB, less than 25.41% of C++ online submissions for Sort Array By Parity.

class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& nums) {
        int p = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] % 2 == 0) {
                swap(nums[p++], nums[i]);
            }
        }
        return nums;
    }
};

익명
작성일 2022.05.03 00:45

class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& nums) {
        for(auto [slow, fast] = tuple(nums.begin(), nums.begin()); fast != nums.end(); ++fast) {
            if (!(*fast%2)) swap(*(slow++), *fast);
        }
        return nums;
    }
};