<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 {
private:
    void recursivepermute(int index, vector < int > & nums, vector < vector < int >> & ans) {
      if (index == nums.size()) {
        ans.push_back(nums);
        return;
      }
      for (int i = index; i < nums.size(); i++) {
        swap(nums[index], nums[i]);
        recursivepermute(index + 1, nums, ans);
        swap(nums[index], nums[i]);
      }
    }
public:
    vector < vector < int >> permute(vector < int > & nums) {
      vector < vector < int >> ans;
      recursivepermute(0, nums, ans);
      return ans;
    }
};
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