<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:
    void reverse(vector<int>&nums,int i,int j)
    {
        while(i<j){
            int t=nums[i];
            nums[i]=nums[j];
            nums[j]=t;
            j--;
            i++;
        }
    }
    vector<int> nextperm(vector<int>& nums) {
        int n=nums.size()-1,i,b=-1;
        for(i=n;i>0;i--){
            if(nums[i]>nums[i-1]){
                 b=i-1;
                 break;
            }
        }
        if (b==-1){ 
            reverse(nums,0,n);
            return nums;
        }
        for(int j=n;j>b;j--){
            if(nums[j]> nums[b]) {
                int t=nums[b];
                nums[b]=nums[j];
                nums[j]=t;
                break;
            }
        }
        reverse(nums,b+1,n);
        return nums;
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> temp=nums;
        res.push_back(temp);
        nextperm(temp);
        while(temp!=nums){
            res.push_back(temp);
            nextperm(temp);            
        }
        return res;
    }
};
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