<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:
    void permute(vector<int>& nums, unordered_set<int>& visited,
                 vector<int>& currentPermutation,
                 vector<vector<int>>& permutations) {
        if (visited.size() == nums.size()) {
            permutations.emplace_back(currentPermutation);
            return;
        }

        for (auto i : nums) {
            if (visited.contains(i)) {
                continue;
            }
            visited.insert(i);
            currentPermutation.push_back(i);
            permute(nums, visited, currentPermutation, permutations);
            currentPermutation.pop_back();
            visited.erase(i);
        }
    }

    vector<vector<int>> permute(vector<int>& nums) {
        unordered_set<int> visited;
        vector<vector<int>> returnValue;
        vector<int> currentPermutation;
        permute(nums, visited, currentPermutation, returnValue);
        return returnValue;
    }
};
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