Print all the permutations of a string in C

Question

I am learning backtracking and recursion and I am stuck at an algorithm for printing all the permutations of a string. I solved it using the bell algorithm for permutation b

Answer 1

I create more specific but not efficient Program for permutation for general string.
It's work nice way.
//ubuntu 13.10 and g++ compiler but it's works on any platform and OS
//All Permutation of general string.

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
using namespace std;
int len;
string str;

void permutation(int cnum)
{
    int mid;
    int flag=1;
    int giga=0;
    int dead=0;
    int array[50];
    for(int i=0;i<len-1;i++)
    {
        array[50]='\0';
        dead=0;
        for(int j=cnum;j<len+cnum;j++)
        {
            mid=j%len;
            if(mid==cnum && flag==1)
            {
                cout<<str[mid];
                array[dead]=mid;
                dead++;
                flag=0;
            }
                else
            {
                giga=(i+j)%len;
                for(int k=0;k<dead;k++)                 
                {
                    if((array[k]==giga) && flag==0)
                    {
                    giga=(giga+1)%len;
                    }
                }
                cout<<str[giga];
                array[dead]=giga;
                dead++;

            }
        }
        cout<<endl;
        flag=1; 
    }
}
int main()
{
    cout<<"Enter the string :: ";
    getline(cin,str);
    len=str.length();
    cout<<"String length = "<<len<<endl;
    cout<<"Total permutation = "<<len*(len-1)<<endl;
    for(int j=0;j<len;j++)
    {
        permutation(j);
    }
return 0;
}

Answer 2

The code has 2 problems, both related to n, the assumed length of the string. The code for (j = i; j <= n; j++) { swap((a+i), (a+j)); ... swap in string's null character '\0' and gives code truncated results. Check the original (i == n) which should be (i == (n-1)).

Backtracking is applied by calling swap() twice effective undoing its original swap.

The order of complexity is the same for Bell Algorithm.

#include <stdio.h>

void swap(char *a, char *b) { char t = *a; *a = *b; *b = t; }

void permute(char *a, int i, int n) {
   // If we are at the last letter, print it
   if (i == (n-1)) printf("%s\n", a);
   else {
     // Show all the permutations with the first i-1 letters fixed and 
     // swapping the i'th letter for each of the remaining ones.
     for (int j = i; j < n; j++) {
       swap((a+i), (a+j));
       permute(a, i+1, n);
       swap((a+i), (a+j));
     }
  }
}

char s[100];
strcpy(s, "ABCD");
permute(s, 0, strlen(s));