Sort Array By Parity
<p>Given an integer array <code>nums</code>, move all the even integers at the beginning of the array followed by all the odd integers.</p>

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

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

<pre>
<strong>Input:</strong> nums = [3,1,2,4]
<strong>Output:</strong> [2,4,3,1]
<strong>Explanation:</strong> The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0]
<strong>Output:</strong> [0]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 5000</code></li>
</ul>
Array; Two Pointers; Sorting
function sortArrayByParity(nums: number[]): number[] {
    for (let i = 0, j = nums.length - 1; i < j; ) {
        if (nums[i] % 2 === 0) {
            ++i;
        } else if (nums[j] % 2 === 1) {
            --j;
        } else {
            [nums[i], nums[j]] = [nums[j], nums[i]];
            ++i;
            --j;
        }
    }
    return nums;
}
