<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 {
  void dfs(vector<int> &nums, vector<int> &indices,
           vector<vector<int>> &result) {
    if (indices.size() >= nums.size()) {
      vector<int> v;
      for (auto i : indices) {
        v.push_back(nums[i]);
      }
      if (find(result.begin(), result.end(), v) == result.end())
        result.push_back(v);
      return;
    }
    for (int i = 0; i < nums.size(); ++i) {
      if (find(indices.begin(), indices.end(), i) != indices.end())
        continue;
      indices.push_back(i);
      dfs(nums, indices, result);
      indices.pop_back();
    }
  }

public:
  vector<vector<int>> permuteUnique(vector<int> &nums) {
    vector<int> indices;
    vector<vector<int>> result;
    dfs(nums, indices, result);
    return result;
  }
};
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