Sort Array By Parity II

Description

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Example 2:

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

Constraints:

2 <= nums.length <= 2 * 10⁴
nums.length is even.
Half of the integers in nums are even.
0 <= nums[i] <= 1000

Follow Up: Could you solve it in-place?

Solution(javascript)

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortArrayByParityII = function(nums) {
    let even = 0
    let odd = 1
    
    while (even < nums.length && odd < nums.length){
        if (nums[even] % 2 === 0) {
            even += 2
        } else if (nums[odd] % 2 === 1) {
            odd += 2
        } else {
            let temp = nums[odd]
            nums[odd] = nums[even]
            nums[even] = temp
        }
    }
    return nums
};