Permutations II
<p>Given a collection of numbers, <code>nums</code>,&nbsp;that might contain duplicates, return <em>all possible unique permutations <strong>in any order</strong>.</em></p>

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

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

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

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 8</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
</ul>
Array; Backtracking; Sorting
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permuteUnique = function (nums) {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const ans = [];
    const t = Array(n);
    const vis = Array(n).fill(false);
    const dfs = i => {
        if (i === n) {
            ans.push(t.slice());
            return;
        }
        for (let j = 0; j < n; ++j) {
            if (vis[j] || (j > 0 && nums[j] === nums[j - 1] && !vis[j - 1])) {
                continue;
            }
            t[i] = nums[j];
            vis[j] = true;
            dfs(i + 1);
            vis[j] = false;
        }
    };
    dfs(0);
    return ans;
};
