Sort Array By Parity II - LeetCode solutions

Problem

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 * 10^4
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 (Java)

class Solution {
    public int[] sortArrayByParityII(int[] nums) {
        int i = 0;
        int j = 1;
        while (i < nums.length - 1 && j < nums.length) {
            if (nums[i] % 2 != 0 && nums[j] % 2 == 0) {
                int tmp = nums[i];
                nums[i] = nums[j];
                nums[j] = tmp;
                i += 2;
                j += 2;
            }
            while (i < nums.length - 1 && nums[i] % 2 == 0) {
                i += 2;
            }
            while (j < nums.length && nums[j] % 2 != 0) {
                j += 2;
            }
        }
        return nums;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).