<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) {
        if(nums.size() == 1) {
            vector<vector<int>> result;
            vector<int> tmp{nums[0]};
            result.push_back(tmp);
            return result;
        }

        vector<vector<int>> totalResult;
        for(int i = 0; i < nums.size(); i++) {
            int last = nums[i];
            vector<int> interNums = nums;
            interNums.erase(interNums.begin()+i);
            vector<vector<int>> result = permute(interNums);
            for(int j = 0; j < result.size(); j++) {
                result[j].push_back(last);
            }

            for(int j = 0; j < result.size(); j++) {
                totalResult.push_back(result[j]);
            }            
        }

        return totalResult;        
    }
};
12300