Find if digits in two numbers are permutations of each other

I was working on Project Euler problem 72

One subtask was to create function to check where numbers "a" and "b" are permutations of each other...
When I finish this problem and check other solutions on forum...I found few C++ solutions. The realization of this function was awful..and extremely slow.

Here is mine:

bool is_permutation(int a,int b)

{

    char test1[10]={0};

    char test2[10]={0};

    while(a>0 && b>0)

    {

        ++test1[a%10];

        ++test2[b%10];

        a/=10;b/=10;

    }

    if(a!=b)

        return false;

    return memcmp(test1,test2,10)==0?true:false;

}

There was one really close realization...
(you can ++test1[a%10];--test2[b%10]; and at the end check if test1=={0} )

But the others...

1st (O(n^2) on sorting numbers... cmon man can't you just qsort then):

#define CAP_LENGTH 8

inline int permutation(int a, int b){

    static int a_digits[CAP_LENGTH];

    static int b_digits[CAP_LENGTH];

    static int i,j, a_length, b_length;

    i=0;

    do{

        a_digits[i]=a%10;

        i++;

    } while( (a/=10)!=0 );

    a_length=i;

    i=0;

    do{

        b_digits[i]=b%10;

        i++;

    } while( (b/=10)!=0 );

    if(a_length!=i)

        return 0; //even digit number differ between a and b

    b_length=i;

    for(i=0; i<a_length; i++){

        // cycle on all a digits

        for (j=0; j<b_length; j++){

            // cycle on all the (remaining, valid) b digits

            if(a_digits[i]==b_digits[j]){

                b_digits[j]=-1;

                break;

            }

        }

        if(j==b_length)

            return 0;

    }

    return 1;

}

2nd (gogo STL...):

bool are_permutations (int num1, int num2)

{

    /* Must have the same number of digits. */

    if ((int) log10 (num1) != (int) log10 (num2))

        return false;

 

    std::deque<char> digits1, digits2;

    int sorted1 = 0, sorted2 = 0;

 

    while (num1 > 0)

    {

        digits1.push_back (num1 % 10);

        num1 *= 0.1;

    }

 

    while (num2 > 0)

    {

        digits2.push_back (num2 % 10);

        num2 *= 0.1;

    }

 

    std::sort (digits1.begin (), digits1.end ());

    std::sort (digits2.begin (), digits2.end ());

 

    for (std::deque<char>::iterator i = digits1.begin (); i != digits1.end ();

        ++i)

        sorted1 = sorted1 * 10 + *i;

 

    for (std::deque<char>::iterator i = digits2.begin (); i != digits2.end ();

        ++i)

        sorted2 = sorted2 * 10 + *i;

 

    return sorted1 == sorted2;

}

3rd (and the winner is...sprintf, qsort,strlen...):

bool perm(int a, int b)

{

    static char b1[1024];

    static char b2[1024];

 

    sprintf(b1, "%d", a);

    sprintf(b2, "%d", b);

 

    int l1 = strlen(b1);

    int l2 = strlen(b2);

 

    if (l1 != l2)

        return false;

 

    qsort(b1, l1, 1, cmp);

    qsort(b2, l2, 1, cmp);

 

    return !strcmp(b1, b2);

}