<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>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());

        return per(nums);  
    }

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

        vector<vector<int>> totalResult;
        for(int i = 0; i < nums.size(); i++) {
            if(i != 0 && nums[i] == nums[i-1]) {
                continue;
            }

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

        return totalResult;  
    }    
};
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