Given a sequence that can contain repeated numbers , Returns all non repeating full permutations .

Examples :
input : [1,1,2] output : [ [1,1,2], [1,2,1], [2,1,1] ]

Thinking analysis : The difference between this question and the last one is that there are repeated elements in the array , Therefore, the emphasis of this topic is to remove the duplicate . If you have no idea about this topic , There are many ways to do it online , You can browse them all , Find the best way for yourself .

use dfs The code is as follows :

class Solution {
    vector<vector<int>>result;// For holding, so the answer
    void dfs(vector<int>& nums,vector<int>& temp,vector<bool>& sign)
        if(temp.size()==nums.size())// If temp I put it in nums All elements in , It means that we have found an answer that satisfies the conditions
            for(int i=0;i<nums.size();i++)
                 if(sign[i]==false)// If it's been used , skip

// If you repeat the number in front of him , And the number in front of you should also participate in the arrangement , Because there is only one arrangement for the two repeated numbers , The first time the recursion reaches the end point, it will be liberated to the solution set , So skip

            temp.push_back(nums[i]);// New elements put temp in , Construct solution
            sign[i]=false;// Indicates that it has been used
            dfs(nums,temp,sign); Recursion continues to find the elements that make up the solution

// There are no elements available to make up the solution , After getting a solution , Look back to see if there are any unsolved solutions in the previous step


    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<int> temp;// Temporary array , A solution satisfying a condition
        vector<bool>sign(nums.size(),true);// Used to mark whether this has been used
        sort(nums.begin(),nums.end());// Sorting makes it easy for us to remove the duplicate
        return result;

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