# 922. Sort Array By Parity II (Easy)

https://leetcode.com/problems/sort-array-by-parity-ii/

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`

## Solutions

``````class Solution {
private void swap(int[] A, int i, int j) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}

public int[] sortArrayByParityII(int[] A) {
int size = A.length;

// stands for the even index pointer
int evenPtr = 0;

// stands for the odd index pointer
int oddPtr = 1;

for (int i = 0; i < size; i++) {
// find odd value at even index
while (evenPtr < size) {
if (A[evenPtr] % 2 != 0) {
break;
}

evenPtr += 2;
}

// find even value at odd index
while (oddPtr < size) {
if (A[oddPtr] % 2 != 1) {
break;
}

oddPtr += 2;
}

if (evenPtr < size && oddPtr < size) {
swap(A, evenPtr, oddPtr);
}
}

return A;
}
}
``````