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// Source : https://oj.leetcode.com/problems/permutations-ii/

// Author : Hao Chen

// Date : 2014-06-21

/**********************************************************************************

*

* 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].

*

*

**********************************************************************************/

#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;

}