如何使用递归查找数字中的最小元素 [C]

Question

ok so I am preparing for my C exams and I'm a bit stuck when it comes to recursions I'm a freshman at my university and this seems a bit difficult to me, the exercise requires that in a given number using a recursive function I need to find the smallest element, ex:52873 would be 2 and the program would need to print 2.

#include <stdio.h>

    int min (int number, int element){
        if (number==0)
            return element;
        if (element>number%10)
            element=number%10;

        min(number/10,element);
    }

    int main (){
        int number;
        while (scanf("%d",&number)){
            printf("%d\n",min(number,9));

        }
    }

this is the code for the answer of the exercise but I do not understand it and would love to get some insight on why it is solved like this since I don't really understand it and different ways to solve it, thank you so much in advance.

Answer 1

The idea of the program is the following:

when you % a number by 10, you get the last digit, ex:

16%10 = 6;
6%10 = 6;
536%10 = 6;

When you / a number by 10, you take out of the number the last digit, ex:

16/10 = 1;
6/10 = 0; (6 = 06)
536/10 = 53;

So, the idea of your recursive answer is save in "element" the smallest digit, and analyse each digit recursively by dividing the number by 10 (/) and doing module (%).

But, i think your solution have a problem, if my number is 0, you'll return 9 as answer.

Answer 2

For More Clearity Let's See a Example

How recursion works

let's element=9

it every time check last digit

min(9154,element)                       -----first call
if(element>4)       
    element=4       -------//now element is 4               
        min(915,element)    -------9154/10=  915        ------second call

        if(element>5)   ------here element is 4 so  no update

                min(91,element)  ------915/10=91        ------third call

                if(element>1)
                      element=1   ---------now element is 1
                       min(9,element)       ---------91/10=9    -----fourth call

                       if(9>element)     ---------no update 
                            min(0,element)   ------9/10=0  
                        //since become 0 so it return element

Answer 3

int min (int number, int element){
    if (number==0)
        return element;
    if (element>number%10)
        element=number%10;

    min(number/10,element);
}

Here, your recursive function has two arguments. int number which stores the number of which the least element is to be obtained and int element to store the least element.

In each call, you check the following two conditions:

• if (number > 0) then return element .
• if (element > number%10) . Here, number % 10 denotes the last number of the number and it's compared against element which is initially set to 9 . If the number % 10 is less than element , then the value of the element is set to number % 10 .

After the above process, you recursively call the function with the values number / 10 and element . The reason why you call number \\ 10 is because it removes the last digit from the number which was already checked for in the previous call.

And finally, this process goes on till you have no digits are left ie, number becomes zero.

Example: take min(1234, 9) as the call from main() function.

1. call from main() function

    number = 1234 element = 9 since number != 0 continue; since, 1234 % 10 ie, 4 < 9, element is set to 4 

   and the function is recursively called with values number = 1234 / 10 = 123 and element = 4

2. first recursive function call:

    number = 123 element = 4 since number != 0 continue; since, 123 % 10 ie, 3 < 4, element is set to 3 

   and the function is recursively called with values number = 123 / 10 = 12 and element = 3

3. second recursive function call:

    number = 12 element = 3 since number != 0 continue;. since, 12 % 10 ie, 2 < 3, element is set to 2 and the function is recursively called with values `number = 12 / 10 = 1` and `element = 2` 

4. third recursive function call:

    number = 1 element = 2 since number != 0 continue; since, 1 % 10 ie, 1 < 2, element is set to 1 

   and the function is recursively called with values number = 1 / 10 = 0 and element = 1

5. finally in the fourth and last recursive call:

    number = 0 element = 1 since `number = 0`, the value of the `element` ie, `1` is returned. 

Answer 4

find the lowest digit of an integer > 0 using recursion:

int min_digit_rec(int num)
{
if (num < 10)
    return num;
else    
        {
            int l_digit;
            int r = min_digit_rec(num / 10);
            
            l_digit = num % 10;
            return r < l_digit ? r : l_digit;
        }
    
}