905. Sort array by parity

May 2, 2022

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

code

class Solution {
    public int[] sortArrayByParity(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left < right){
            while (left < nums.length && (nums[left] & 1) != 1) left++;
            while (right >= 0 && (nums[right] & 1) == 1) right--;
            if (left < right){
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
            }
        }
        return nums;
    }
}

code

class Solution:
    def sortArrayByParity(self, A: List[int]) -> List[int]:
        start, end = 0, len(A) - 1
        while start < end:
            while A[start] % 2 == 0 and start < end:
                start += 1
            while A[end] % 2 == 1 and start < end:
                end -= 1
            A[start], A[end] = A[end], A[start]
            start += 1
            end -= 1
        return A