<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) {

    vector<vector<int>> ans;

    dfs(nums, vector<bool>(nums.size()), {}, ans);

    return ans;

  }

 private:

  void dfs(const vector<int>& nums, vector<bool>&& used, vector<int>&& path,

           vector<vector<int>>& ans) {

    if (path.size() == nums.size()) {

      ans.push_back(path);

      return;

    }

    for (int i = 0; i < nums.size(); ++i) {

      if (used[i])

        continue;

      used[i] = true;

      path.push_back(nums[i]);

      dfs(nums, std::move(used), std::move(path), ans);

      path.pop_back();

      used[i] = false;

    }

  }

};


       
    

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