<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 backtrack(vector<vector<int>>& resultList, vector<int>& tempList, vector<int>& nums) {
        // If we match the length, it is a permutation
        if (tempList.size() == nums.size()) {
            resultList.push_back(tempList);
            return;
        }

        for (int number : nums) {
            // Skip if we get the same element
            if (find(tempList.begin(), tempList.end(), number) != tempList.end())
                continue;

            // Add the new element
            tempList.push_back(number);

            // Go back to try other elements
            backtrack(resultList, tempList, nums);

            // Remove the element
            tempList.pop_back();
        }
    }
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> resultList;
        vector<int> tempList;
        backtrack(resultList, tempList, nums);
        return resultList;
    }
};
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