(Easy) Sort Array By Parity II - LeetCode

Description:

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

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

You may return any answer array that satisfies this condition.

 

Example 1:

Input: [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.

 

Note:

1. 2 <= A.length <= 20000
2. A.length % 2 == 0
3. 0 <= A[i] <= 1000

 

Accepted

55,816

Submissions

83,110

Solution:

class Solution {
    public int[] sortArrayByParityII(int[] A) {
        
        if(A==null||A.length ==0){
            return null;
        }
        
        int[] B = new int[A.length];
        
        int k = 0;
        int j =1; 
        for(int i = 0; i<A.length; i++ ){
            
            
            if(A[i]%2 ==0){
                B[k] = A[i];
                
                k = k+2; 
            }
            
            else{
                
                B[j] = A[i];
                
                j = j+2; 
                
            }
            
        }
        
        return B;
    }
}