LeetCode-in-Java

922. Sort Array By Parity II

Easy

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

public 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;
    }
}

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