<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:
    unordered_set<int> record;
    void DFS(vector<int>& nums, vector<vector<int>>& res, vector<int> vec)
    {
        if (record.size() == nums.size())
        {
            res.push_back(vec);
            return;
        }

        for (int i=0;i<nums.size();i++)
        {
            if (record.count(i))
                continue;

            record.insert(i);
            vec.push_back(nums[i]);

            DFS(nums, res, vec);
            
            vec.pop_back();
            record.erase(i);
            
        }

    }
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> res;
        DFS(nums, res, vector<int>());
        return res;
        
    }
};
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