<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) {
        if (nums.size() == 1)
            return vector<vector<int>>{nums};
        vector<vector<int>> store{};
        vector<int> cur_step{};
        cur_step.reserve(nums.size());
        perm(store, nums, 0, cur_step);

        sort(store.begin(), store.end());
        auto iter = unique(store.begin(), store.end());
        return vector<vector<int>>(store.begin(), iter);
    }

    void perm(vector<vector<int>> &store, vector<int> &nums, size_t cur_index, vector<int> &cur_step)
    {
        if (nums.size() - cur_index == 1)
        {
            cur_step.push_back(nums[cur_index]);
            store.push_back((cur_step));
            cur_step.pop_back();
            return ;
        }
        else
        {
            for (size_t index{cur_index}; index < nums.size(); ++index)
            {
                swap(nums[cur_index], nums[index]);
                cur_step.push_back(nums[cur_index]);
                perm(store, nums, cur_index+1, cur_step);
                swap(nums[cur_index], nums[index]);
                cur_step.pop_back();
            }
        }
    }
};
35701