Print all permutation in lexicographic order

Question

I want to print all permutation of string in lexicographic order. I wrote this code:

void permute(char *a, int i, int n) {
   if (i == (n-1)) printf(\"\\\"%s         


        

Answer 1

IMHO, it would be simpler to first sort the characters of the string because the number of permutations (n!) is always greater (or equal for n = 1 or 2) than the number of characters.

And you were not far from the solution but you must rotate instead of swap. Here is a slight variation that returns an array of dynamically allocated strings that are printed in the main :

#include 
#include 
#include 

int char_compare(const void *a, const void *b) {
    return (*((char *) a)) - (*((char *) b));
}

int fact(int n) {
    int f = 1;
    while (n > 0) {
        f *= n--;
        if (f < 0) return 0;
    }
    return f;
}

void rotateback(char *a, int i, int j) {
    int k;
    char tmp;
    tmp = a[i];
    for(k=i; ki; k--) {
        a[k] = a[k-1];
    }
    a[i] = tmp;
}

void permute(char *a, int i, int n, char ***permuts) {
    int j;
    if (i == (n-1)) {
        **permuts = strdup(a);  // store a copy of the string
        *permuts += 1;          // and point to next location
    }
   else {
       for (j = i; j < n; j++) {
           rotate(a, i, j);
           permute(a, i+1, n, permuts);
           rotateback(a, i, j);
       }
   }
}
char ** permutations(const char *str_orig) {
    int i, j;
    size_t n = strlen(str_orig);
    size_t fact_n = fact(n);
    char ** permuts, **work;

    char * str = strdup(str_orig); // get a local copy
    qsort(str, n, 1, char_compare); // and sort it

    permuts = work = calloc(fact_n, sizeof(char *)); // allocate n! pointers

    permute(str, 0, n, &work);
    return permuts;
}

int main() {
    char str[] = "cab";
    int i;

    char **permuts = permutations(str);

    for (i=0; i

Output is correctly :

"abc"
"acb"
"bac"
"bca"
"cab"
"cba"