<p>Given a collection of numbers, <code>nums</code>,&nbsp;that might contain duplicates, return <em>all possible unique permutations <strong>in any order</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,2]
<strong>Output:</strong>
[[1,1,2],
 [1,2,1],
 [2,1,1]]
</pre>

<p><strong class="example">Example 2:</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>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 8</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> sol;  
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<int> temp;  
        vector<bool> check(nums.size());
        sort(nums.begin(),nums.end());
        backtrack(temp, nums, check);
        return sol;
    }

private:
    void backtrack(vector<int>& temp, vector<int> nums, vector<bool>& check) {
        if (temp.size() == nums.size() && find(sol.begin(), sol.end(), temp)==sol.end()) {
            sol.push_back(temp);  
            return;
        }
        for (int i=0;i<nums.size();i++) {
            if (check[i]==true)
                continue; 
            // add the element
            check[i]=true;
            temp.push_back(nums[i]);
            // go back
            backtrack(temp, nums, check);
            // remove the last element
            check[i]=false;
            temp.pop_back();
        }
    }
};
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