LeetCode-Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

第一个版本，加了一个set去除重复

class Solution {
public:
    void sub(vector<vector<int> > & ret,vector<int>& num,int index){
        if(index==num.size()){
            ret.push_back(num);
            return ;
        }
        int temp;
        sub(ret,num,index+1);
        for(int i=index+1;i<num.size();i++){
            if(num[i]!=num[index]){
                temp=num[i];
                num[i]=num[index];
                num[index]=temp;
                sub(ret,num,index+1);
                temp=num[i];
                num[i]=num[index];
                num[index]=temp;
            }
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<vector<int> >ret;
        sub(ret,num,0);
        set<vector<int> > sec(ret.begin(),ret.end());
        vector<vector<int> >ret2(sec.begin(),sec.end());
        return ret2;
    }
};

第二个版本，改变策略直接填入

class Solution {
public:
    void sub(vector<vector<int> > &ret,vector<int>& num,vector<int>&num2,vector<int>&count,vector<int>& count2,int index){
        if(index==num.size()){
            ret.push_back(num);
            return ;
        }
        for(int i=0;i<num2.size();i++){
            if(count2[i]<count[i]){
                num[index]=num2[i];
                count2[i]++;
                sub(ret,num,num2,count,count2,index+1);
                count2[i]--;
            }
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<int>num2;
        vector<int>count;
        sort(num.begin(),num.end());
        if(num.size()==0)return vector<vector<int> >();
        num2.push_back(num[0]);
        count.push_back(1);
        for(int i=1;i<num.size();i++){
            if(num[i]!=num[i-1]){
                num2.push_back(num[i]);
                count.push_back(1);
            }
            else{
                count[count.size()-1]++;
            }
        }
        vector<int>count2;
        count2.resize(count.size(),0);
        vector<vector<int> >ret;
        sub(ret,num,num2,count,count2,0);
        return ret;
    }
};