<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:
    vector<vector<int>> permute(vector<int>& nums) {

        if ( nums.empty() )
            return {};

        if ( nums.size() == 1 ){
            return { {nums[0]} };
        }

        vector<vector<int>> output, returned;

        for ( int i = nums.size()-1; i >= 0; i-- ){
            vector<int> subset;
            for ( int j = 0; j < nums.size(); j++ ){
                if ( j != i ){
                    subset.push_back(nums[j]);
                }
            }
            returned = permute(subset);

            for ( int j = 0; j < returned.size(); j++ ){
                returned[j].push_back(nums[i]);
                output.push_back(returned[j]);
            }
        }
        return output;
    }
};
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