3Sum
<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>

<p>Notice that the solution set must not contain duplicate triplets.</p>

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

<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong> 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>

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

<ul>
	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Two Pointers; Sorting
class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer[][]
     */
    function threeSum($nums) {
        sort($nums);
        $ans = [];
        $n = count($nums);
        for ($i = 0; $i < $n - 2 && $nums[$i] <= 0; ++$i) {
            if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
                continue;
            }
            $j = $i + 1;
            $k = $n - 1;
            while ($j < $k) {
                $x = $nums[$i] + $nums[$j] + $nums[$k];
                if ($x < 0) {
                    ++$j;
                } elseif ($x > 0) {
                    --$k;
                } else {
                    $ans[] = [$nums[$i], $nums[$j++], $nums[$k--]];
                    while ($j < $k && $nums[$j] == $nums[$j - 1]) {
                        ++$j;
                    }
                    while ($j < $k && $nums[$k] == $nums[$k + 1]) {
                        --$k;
                    }
                }
            }
        }
        return $ans;
    }
}