<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> > ans;
    void Rec(vector<int> & candidates,set<int> &visited,vector<int>& nums)
    {
        if(candidates.size()==nums.size())
        {
            ans.push_back(candidates);
            return;
        }
        for(int i=0;i<nums.size();i++)
        {
            if(visited.find(i)==visited.end())
            {
                candidates.push_back(nums[i]);
                visited.insert(i);
                Rec(candidates,visited,nums);
                candidates.pop_back();
                visited.erase(i);
            }
        }
    }
    vector<vector<int>> permute(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<int> candidates;
        set<int> visited;
        Rec(candidates,visited,nums);
        set<vector<int>> s=set(ans.begin(),ans.end());
        vector<vector<int>> answer;
        for(auto it=s.begin();it!=s.end();it++)
        {
            answer.push_back(*it);
        }
        return answer;
    }
};
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