Permutations
<p>Given an array <code>nums</code> of distinct integers, return all the possible <span data-keyword="permutation-array">permutations</span>. You can 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 = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> [[0,1],[1,0]]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1]
<strong>Output:</strong> [[1]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 6</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>All the integers of <code>nums</code> are <strong>unique</strong>.</li>
</ul>
Array; Backtracking
func permute(nums []int) (ans [][]int) {
	n := len(nums)
	t := make([]int, n)
	vis := make([]bool, n)
	var dfs func(int)
	dfs = func(i int) {
		if i == n {
			ans = append(ans, slices.Clone(t))
			return
		}
		for j, x := range nums {
			if !vis[j] {
				vis[j] = true
				t[i] = x
				dfs(i + 1)
				vis[j] = false
			}
		}
	}
	dfs(0)
	return
}
