Sort Array By Parity II
<p>Given an array of integers <code>nums</code>, half of the integers in <code>nums</code> are <strong>odd</strong>, and the other half are <strong>even</strong>.</p>

<p>Sort the array so that whenever <code>nums[i]</code> is odd, <code>i</code> is <strong>odd</strong>, and whenever <code>nums[i]</code> is even, <code>i</code> is <strong>even</strong>.</p>

<p>Return <em>any answer array that satisfies this condition</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,2,5,7]
<strong>Output:</strong> [4,5,2,7]
<strong>Explanation:</strong> [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3]
<strong>Output:</strong> [2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>nums.length</code> is even.</li>
	<li>Half of the integers in <code>nums</code> are even.</li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow Up:</strong> Could you solve it in-place?</p>
Array; Two Pointers; Sorting
function sortArrayByParityII(nums: number[]): number[] {
    for (let i = 0, j = 1; i < nums.length; i += 2) {
        if (nums[i] % 2) {
            while (nums[j] % 2) {
                j += 2;
            }
            [nums[i], nums[j]] = [nums[j], nums[i]];
        }
    }
    return nums;
}
