Sort Array By Parity II - LeetCode

 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

Solution(Only function):

class Solution {

    public int[] sortArrayByParityII(int[] nums) {

        

        int i=0;

        int j=1;

    while(i<nums.length && j<nums.length){    

        while(i<nums.length)

        {

            if(nums[i]%2!=0)

                break;

            

            i+=2;

        }

        while(j<nums.length)

        {

            if(nums[j]%2==0)

                break;

            

            j+=2;

        }

       

      if(i<nums.length && j<nums.length){  

        int temp=nums[i];

        nums[i]=nums[j];

        nums[j]=temp;

        }

    }    

    return nums;    

    }       

}