How to get an integral reverse after a long integer

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How to get a long integer integer after reverse

Recently I encountered such a problem in writing code: How to get to get a long integer after a long integer number in reverse order? Such as input \ (A = 12345678 \) , to obtain an output \ (RA = 87654321 \) .

Below, a careful analysis of the characteristics of the number of input, you can get a simple recursive algorithm to solve this problem.

Note: The final countdown time to make plans less the second line plus

void reverse(unsigned long a, unsigned long *sum, unsigned long power)
{
    if (a > 10)
        reverse(a/10, sum, power/10);
    *sum += (a % 10) * power;
}

unsigned long get_reverse(unsigned long a)
{
    /* 根据a的位数确定最大位权值 */
    int i;
    unsigned long power = 1;
    char buf[16];
    sprintf(buf, "%lu", a);
    for (i = 0; i < strlen(buf) - 1; i++)
        power *= 10;

    /* 调用reverse() */
    unsigned long sum = 0;
    reverse(a, &sum, power);
    return sum;
}

Want to come back to the above code, you will find now borrow sprintf () can easily be \ (a \) the maximum bit weight calculated, then why this inefficiency is also recursive calculation with the use of it? Thus, a little better efficiency, avoid recursive algorithm can be implemented as follows.

unsigned long get_reverse(unsigned long a)
{
    /* 根据a的位数确定最大位权值 */
    int i;
    unsigned long power = 1;
    char buf[16];
    sprintf(buf, "%lu", a);
    for (i = 0; i < strlen(buf) - 1; i++)
        power *= 10;

    unsigned long sum = 0;
    for (i = strlen(buf) - 1; i >= 0; i--)
    {
        sum += (buf[i] - '0') * power;
        power /= 10;
    }
    return sum;
}

To the complexity of the code is: \ (O (strlen (A)) = O (log ~ A) \) . Furthermore, there is no more efficient algorithm to solve this problem?