Count Total Digits in a Number useing recursion and when n is 0 then count it's and print 1 . and my question is how to count 0 as digits and print 1?

Saiyed Mohammad Adnan 0
Viewed 26
public class CountNumberRecursion 
{
    
    public static int countDigits(int n)
    {
        int count=0;
        
        if(n==0)
            return n;
        
        return ++count + countDigits(n/10);
    }

    public static void main(String[] args) 
    {
         System.out.println(countDigits(0));
    }

}

In this example run, I want to count 0 as digits and print 1 in output, but it prints 0.

java recursion algorithm
1 Answers

This is an understandable issue. On the one hand we have the case where a single digit is passed as n, and then the recursive call for n/10 should return 0 (as it does now).

On the other hand, if the value n is 0 from the start, it should return 1.

This is a contradiction that you can solve by making 0 a special case in the main program. But this is solved more elegantly by stopping the recursion one step earlier, so not when all digits are gone, but when there is one digit. This you can implement by changing:

if(n==0)
    return n;

...to:

if(n<10)  // Only 1 digit?
    return 1;

This way the function can never return 0, so even for 0 it will return 1.

Remark

It is bit overkill to have the count variable. You can just add 1 to what the recursive call returns:

return 1 + countDigits(n/10);

With a conditional operator, both the base case and the recursive case can be combined in one expression:

public static int countDigits(int n)
{
    return n < 10 ? 1 : 1 + countDigits(n/10);
}
trincot0
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