<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>> result;
    vector<vector<int>> permute(vector<int>& nums) {
        int n = nums.size();
        vector<int> curr;
        recurPermute(curr,nums,n);
        return result;
    }
    
    void recurPermute(vector<int> curr,vector<int> unused,int n){
        if(unused.size()==0||n==0){
            result.push_back(curr);
            return;
        }
        for(int i = 0; i < n;++i){
            int tmp = unused[i];
            unused.erase(unused.begin() + i);
            curr.push_back(tmp);
            recurPermute(curr,unused,n-1);
            curr.pop_back();
            unused.insert(unused.begin()+i,tmp);
        }
        return;
    }
};


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