4Sum
<p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>

<ul>
	<li><code>0 &lt;= a, b, c, d&nbsp;&lt; n</code></li>
	<li><code>a</code>, <code>b</code>, <code>c</code>, and <code>d</code> are <strong>distinct</strong>.</li>
	<li><code>nums[a] + nums[b] + nums[c] + nums[d] == target</code></li>
</ul>

<p>You may return the answer in <strong>any order</strong>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,0,-1,0,-2,2], target = 0
<strong>Output:</strong> [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2], target = 8
<strong>Output:</strong> [[2,2,2,2]]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Two Pointers; Sorting
function fourSum(nums: number[], target: number): number[][] {
    const n = nums.length;
    const ans: number[][] = [];
    if (n < 4) {
        return ans;
    }
    nums.sort((a, b) => a - b);
    for (let i = 0; i < n - 3; ++i) {
        if (i > 0 && nums[i] === nums[i - 1]) {
            continue;
        }
        for (let j = i + 1; j < n - 2; ++j) {
            if (j > i + 1 && nums[j] === nums[j - 1]) {
                continue;
            }
            let [k, l] = [j + 1, n - 1];
            while (k < l) {
                const x = nums[i] + nums[j] + nums[k] + nums[l];
                if (x < target) {
                    ++k;
                } else if (x > target) {
                    --l;
                } else {
                    ans.push([nums[i], nums[j], nums[k++], nums[l--]]);
                    while (k < l && nums[k] === nums[k - 1]) {
                        ++k;
                    }
                    while (k < l && nums[l] === nums[l + 1]) {
                        --l;
                    }
                }
            }
        }
    }
    return ans;
}
