用递归找到一个两位数的数字

Question

I'm trying to code something in C by using recursion.

The user writes two positive numbers of same length and the program gives him a new number, which is composed like this :

new number unity digit = the smallest digit in the second positive number that the user wrote.

new number ten digit = the biggest digit in the first positive number that the user wrote.

Very simple in fact, here is an example :

5642 and 2371

will give us : 61 .

---

I tried something like this :

#include <stdio.h>

int calcPair(int a, int b){

    int number = calcPair(a/10, b/10);
    int digit1 = (number/10);
    int digit2 = number%10;

    if(digit1 < a%10){
        digit1 = a%10;
    }
    if(digit2 > b%10){
        digit2 = b%10;
    }

    return(number);
}

int main()
{
    int a, b, number=0;
    printf("Please enter two positive number of same length:\n");
    scanf("%d", &a);
    scanf("%d", &b);

    calcPair(a, b);
    printf("The two-digit number composed from %d, %d is: %d", a, b, number);
    return 0;
}

BUT the program doesn't run at all.. and closes.

Maybe someone can correct me ? Or helping me finding the mistake .

Thanks by advance.

Answer 1

Your recursion can never end. Consider the following line in calcPair :

int number = calcPair(a/10, b/10);

This statement will always be executed unless you make it conditional, such as:

int number;
if((a != 0) || (b != 0))
    number = calcPair(a/10, b/10);

Eventually, because you're dividing both numbers by 10, this condition will prove FALSE .

Answer 2

Something like this:

int calcPair(int a, int b){
    int number;
    if (a < 10 && b < 10) {
        number = a*10 + b;
    } else {
        int digita = a%10;
        int digitb = b%10;
        number = calcPair(a/10, b/10);     

        if(digita > number/10){
            number = digita*10 + number%10;
        }

        if(digitb < number%10){
            number = (number/10)*10 + digitb;
        }
    }
    return number;
}

Also, a small fix to the main:

int main()
{
    int a, b, number=0;
    printf("Please enter two positive number of same length:\n");
    scanf("%d", &a);
    scanf("%d", &b);

    number = calcPair(a, b);
    printf("The two-digit number composed from %d, %d is: %d", a, b, number);
    return 0;
}

Answer 3

Whether or not you are allowed to (you did not specify in OP),
here is a recursive search method using strings :

Strings are just an array of char . because you are interested in distinguishing the individual digits within a larger integer, the char data type will be a sufficient size container to facilitate the comparison.
Using arrays of char (strings) within a recursive function with exit criteria of strlen() > 0 will allow you to walk through each integer , and select the appropriate value (min or max).
This approach uses two recursive functions : getMinDigit() and getMaxDigit() , both returning a char representing the maximum value digit, or minimum value digit of their respective original multi-digit integer. These results are then concatenated, and converted back into a two digit integer.

Here is the example code that given:

5642 and 2371

will give us : 61.

char getMinDigit(char *digit)
{
    static char val='9';//largest single digit base 10
    int len=0;
    if(strlen(digit) > 0)
    {
        len = strlen(digit);
        if(digit[len-1] < val) //test for smallest char in string
        {
            val = digit[len-1]; 
            digit[len-1] = 0;
            getMinDigit(digit);
        }
        else 
        {
            digit[len-1] = 0;
            getMinDigit(digit);
        }

    }
    return val; 
}

char getMaxDigit(char *digit)
{
    static char val='0'; //smallest single digit base 10
    int len=0;

    if(strlen(digit) > 0)
    {
        len = strlen(digit);
        if(digit[len-1] > val) //search for largest char in string
        {
            val = digit[len-1];
            digit[len-1] = 0;
            getMaxDigit(digit);

        }
        else 
        {
            digit[len-1] = 0;
            getMaxDigit(digit);
        }
    }
    return val; 

}

int calcPair(int a, int b)
{
    char big[10]={""}, small[10]={""};
    char Big, Small;
    char result[3]={""};
    sprintf(big, "%d", a);
    sprintf(small, "%d", b);
    Big = getMaxDigit(big);  //recursive function
    Small = getMinDigit(small); //recursive function
    sprintf(result, "%c%c", Big, Small);
    return atoi(result);

}



int main(void)
{
    int result = calcPair(5642, 2371);
    printf("%d", result);
    return 0;
//for illustration, hard coded to OP values
        //int a, b, number=0;
        //printf("Please enter two positive number of same length:\n");
        //scanf("%d", &a);
        //scanf("%d", &b);

        //calcPair(a, b);
        //printf("The two-digit number composed from %d, %d is: %d", a, b, number);
        //return 0;



}

Answer 4

I think you can refactor your code to be more expressive of your requirement, with a few helper functions.

int greater(int a, int b)
{
   return (a>b);
}

int less(int a, int b)
{
   return (a<b);
}

int pickDigit(int n, int (*func)(int, int))
{
   int ret = n%10;
   n /= 10;
   while ( n > 0 )
   {
      if ( fun(n%10, ret) )
      {
         ret = n%10;
      }
      n /= 10;
   }
   return ret;
}

int getBiggestDigit(int n)
{
   return pickDigit(n, greater);
}

int gteSmallestDigit(int n)
{
   return pickDigit(n, less);
}

int numDigits(int n)
{
    int ret = 0;
    while (n > 0 )
    {
       ++ret;
       n /= 10;
    }
    return ret;
}

int calcPair(int a, int b)
{

    if ( numDigits(a) != numDigits(b) )
    {
        // Deal with error.
    }

    return betBiggestDigit(a)*10+getSmallestDigit(b);
}