(Js) Leetcode 922. Sort array by parity II

topic:

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

Sort the array so that when A[i] is odd, i is also odd; when A[i] is even, i is also even.

You can return any array that meets the above conditions as the answer.

Example:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4 ,5] will also be accepted.

prompt:

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

Ideas:

Iterate over the array

1) If the element is odd, put it into the corresponding odd position

2) Otherwise, put it in the corresponding even position

Code:

/**
 * @param {number[]} A
 * @return {number[]}
 */
var sortArrayByParityII = function (A) {
    let res = [], even = 0, odd = 1;
    for (let i = 0; i < A.length; i++) {
        if (A[i] & 1) {
            res[odd] = A[i];
            odd += 2;
        } else {
            res[even] = A[i];
            even += 2;
        }
    }
    return res;
};

operation result: