Permutations II Solutions in C++

// Source : https://oj.leetcode.com/problems/permutations-ii/
// Author : Hao Chen
// Date   : 2014-06-21



#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// To deal with the duplication number, we need do those modifications:
//    1) sort the array [pos..n].
//    2) skip the same number.
vector<vector<int> > permute(vector<int> &num) {

    vector<vector<int> > vv;
    vv.push_back(num);

    if (num.size() <2){
        return vv;
    }
        
    int pos=0;
    while(pos<num.size()-1){
        int size = vv.size();
        for(int i=0; i<size; i++){
            //sort the array, so that the same number will be together
            sort(vv[i].begin()+pos, vv[i].end());
            //take each number to the first
            for (int j=pos+1; j<vv[i].size(); j++) {
                vector<int> v = vv[i];
                //skip the same number 
                if (j>0 && v[j]==v[j-1]){
                    continue;
                }
                int t = v[j]; 
                v[j] = v[pos];
                v[pos] = t;
                vv.push_back(v);
            }
        }
        pos++;
    }
    return vv;
}

void printVector( vector<int>&  pt)
{
    cout << "{ ";
    for(int j=0; j<pt.size(); j++){
        cout << pt[j] << " ";
    }
    cout << "} " << endl;
}


int main(int argc, char** argv)
{
    int n = 3;
    if (argc>1){
       n = atoi(argv[1]); 
    }

    srand(time(NULL));
    vector<int> v;
    for (int i=0; i<n; i++) {
        v.push_back(random()%n+1);
    }
    /*v[0] =0;
    v[1] =1;
    v[2] =0;
    v[3] =0;
    v[4] =9;*/

    printVector(v); 
    cout << "-----------------" << endl;


    vector<vector<int> > vv;
    vv = permute(v);
    
    for(int i=0; i<vv.size(); i++) {
        printVector(vv[i]);
    }

    return 0;
}

#include <vector>
using std::vector;
#include <algorithm>
using std::next_permutation; using std::sort;

class Solution {
    public:
        vector<vector<int> > permuteUnique(vector<int> &num) {
            vector<vector<int>> retv;
            sort(num.begin(), num.end());
            do {
                retv.push_back(num);
            } while (next_permutation(num.begin(), num.end()));
            return retv;
        }
};

/// Source : https://leetcode.com/problems/permutations-ii/description/
/// Author : liuyubobobo
/// Time   : 2017-12-18

#include <iostream>
#include <vector>

using namespace std;

/// Most Naive Recursive
/// Time Complexity: O(n^n)
/// Space Complexity: O(n)
class Solution {

private:
    vector<vector<int>> res;
    vector<bool> used;

public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {

        res.clear();
        if(nums.size() == 0)
            return res;

        used = vector<bool>(nums.size(), false);
        sort(nums.begin(), nums.end());
        vector<int> p;

        generatePermutation(nums, 0, p);

        return res;
    }

private:
    void generatePermutation(const vector<int>& nums, int index, vector<int> &p){

        if(index == nums.size()){
            res.push_back(p);
            return;
        }

        for(int i = 0 ; i < nums.size() ; i ++)
            if(!used[i]){
                if(i > 0 && nums[i] == nums[i-1] && !used[i-1])
                    continue;
                p.push_back(nums[i]);
                used[i] = true;

                generatePermutation(nums, index + 1, p);

                p.pop_back();
                used[i] = false;
            }

    }
};


void printVec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    vector<int> nums1 = {1, 1, 2};
    vector<vector<int>> res1 = Solution().permuteUnique(nums1);
    for(const vector<int>& tres: res1)
        printVec(tres);

    vector<int> nums2 = {2, 2, 1, 1};
    vector<vector<int>> res2 = Solution().permuteUnique(nums2);
    for(const vector<int>& tres: res2)
        printVec(tres);

    return 0;
}