<p>Given an array <code>nums</code> of distinct integers, return <em>all the possible permutations</em>. 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>

class Solution {
public:
void permutations_function(int n, vector<int> &nums, vector<int> &ds, vector<int> &visited,
                           vector<vector<int>> &ans)
{
    // Base Condition
    if (ds.size() == n)
    {
        ans.push_back(ds);
        return;
    }

    for (int i = 0; i < n; i++)
    {
        // If this element is not visited, then only take it
        if (visited[i] == 0)
        {
            visited[i] = 1;
            ds.push_back(nums[i]);
            permutations_function(n, nums, ds, visited, ans);

            // Not pop this element out and make visited[i] == 0 again for other permutations
            visited[i] = 0;
            ds.pop_back();
        }
    }
}
vector<vector<int>> permute(vector<int> &nums)
{
    int n = nums.size();
    vector<vector<int>> ans;
    vector<int> ds;
    vector<int> visited(n, 0);
    permutations_function(n, nums, ds, visited, ans);
    return ans;
}
};
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