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
function threeSum(nums: number[]): number[][] {
    nums.sort((a, b) => a - b);
    const ans: number[][] = [];
    const n = nums.length;
    for (let i = 0; i < n - 2 && nums[i] <= 0; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) {
            continue;
        }
        let j = i + 1;
        let k = n - 1;
        while (j < k) {
            const x = nums[i] + nums[j] + nums[k];
            if (x < 0) {
                ++j;
            } else if (x > 0) {
                --k;
            } else {
                ans.push([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;
}
