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
public class Solution {
    public IList<IList<int>> Permute(int[] nums) {
        var ans = new List<IList<int>>();
        var t = new List<int>();
        var vis = new bool[nums.Length];
        dfs(nums, 0, t, vis, ans);
        return ans;
    }

    private void dfs(int[] nums, int i, IList<int> t, bool[] vis, IList<IList<int>> ans) {
        if (i >= nums.Length) {
            ans.Add(new List<int>(t));
            return;
        }
        for (int j = 0; j < nums.Length; ++j) {
            if (!vis[j]) {
                vis[j] = true;
                t.Add(nums[j]);
                dfs(nums, i + 1, t, vis, ans);
                t.RemoveAt(t.Count - 1);
                vis[j] = false;
            }
        }
    }
}
