Tehan Frago

• Feb 13th, 2015

 

Here is an answer that considers all possible permutations including duplicates. If, however, you wanted to remove duplicated, you could save the answers into a vector if they are unique then print all the elements.

Code
#include <string>


#include <iostream>


 


using namespace std;


 


void Permutations(string str)


{


        for(int i = 0; i < str.size(); i++)


        {


                for(int j = 0; j < str.size()-1; j++)


                {


                        char temp = str[j];


                        str[j] = str[j+1];


                        str[j+1] = temp;


                        cout << str << endl;


                }


        }


}


 


int main()


{


        string str;


        cin >> str;


 


        Permutations(str);


 


        return 0;


}


 

Sergey

• May 4th, 2015

 

Hi Tehan! This is a wrong solution. A string of the size N has N! (N factorial) permutations while you have only N*(N-1) ones
One of the solutions is to use some kind of recursion

Rajni kant

• Aug 27th, 2015

 

Code
#include <iostream>


using namespace std;


 


int visited[10000];


char array1[10000];


int count1 = 3;


 


int ttaa;


 


 


void print_perm(char* srt, int index)


{


        if (index == count1) {


                cout << array1<<"


";


                ttaa++;


                return;


        }


        for (int p = 0; p < count1;p++)


        {


                if (visited[p] == 1)continue;


                array1[index] = srt[p];


                visited[p] = 1;


                print_perm(srt,index+1);


                visited[p] = 0;


        }


}


 


void main()


{


        count1 = strlen("abc");


        print_perm("abc", 0);


        


        cout << ttaa << endl;


}

neha

• Aug 27th, 2015

 

Plz see below code

Code
#include <string>


 


 


#include <iostream>


 


 


 


 


 


using namespace std;


 


 


 


 


 


void Permutations(string str)


 


 


{


        for(int i = 0; i < str.size(); i++)


        {


                for(int j = i; j < str.size(); j++)


                {


                        char temp = str[j];


 


 


                        str[j] = str[i];


 


 


                        str[i] = temp;


 


 


                        cout << str << endl;


                }


        }


}


 


 


 


int main()


{


        string str;


        cin >> str;


        Permutations(str);


 


        return 0;


}

Jin Hyuk Cho

• Feb 3rd, 2016

 

A working example

Code
#include <iostream>


#include <string>


#include <set>


#include <sstream>


 


using namespace std;


 


void permute (set<char> &chars, set<string> &permuted) {


    if (chars.size() == 1) {


        stringstream charSS;


        charSS << *(chars.begin());


        permuted.insert(charSS.str());


        return;


    }


    for (set<char>::iterator it = chars.begin(); it != chars.end(); ++it) {


        // constructing remaining characters


        set<char> remainingChars(chars);


        remainingChars.erase(*it);


        // constructing empty set for results


        set<string> permutedRemainingStrings;


        permute(remainingChars, permutedRemainingStrings);


        for (set<string>::iterator rit = permutedRemainingStrings.begin();


             rit != permutedRemainingStrings.end();


             ++rit) {


            stringstream ss;


            ss << *it << *rit;


            string onePermute = ss.str();


            //cout << onePermute << endl;


            permuted.insert(onePermute);


        }


    }


}


int main () {


    


    ostream_iterator<string> output(cout, " " );


    set<char> chars = {c, a, t};


    set<string> permuted;


    permute(chars, permuted);


    copy(permuted.begin(), permuted.end(), output);


    cout << "Done!" << endl;


    


    


    return 0;


}