Squares of a Sorted Array
<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing</strong> order, return <em>an array of <strong>the squares of each number</strong> sorted in non-decreasing order</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [-4,-1,0,3,10]
<strong>Output:</strong> [0,1,9,16,100]
<strong>Explanation:</strong> After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].
</pre>

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

<pre>
<strong>Input:</strong> nums = [-7,-3,2,3,11]
<strong>Output:</strong> [4,9,9,49,121]
</pre>

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

<ul>
	<li><code><span>1 &lt;= nums.length &lt;= </span>10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>nums</code> is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Squaring each element and sorting the new array is very trivial, could you find an <code>O(n)</code> solution using a different approach?
Array; Two Pointers; Sorting
class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer[]
     */
    function sortedSquares($nums) {
        $n = count($nums);
        $ans = array_fill(0, $n, 0);
        for ($i = 0, $j = $n - 1, $k = $n - 1; $i <= $j; --$k) {
            $a = $nums[$i] * $nums[$i];
            $b = $nums[$j] * $nums[$j];
            if ($a > $b) {
                $ans[$k] = $a;
                ++$i;
            } else {
                $ans[$k] = $b;
                --$j;
            }
        }
        return $ans;
    }
}