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javarecursiondigits
Victor Calin
Victor Calin

Reputation: 3

Recursion method that returns number of digits for a given integer

Need to write a method that returns the number of digits foe an integer number.

At first I did it using the iterative approach and everything worked just fine, however, when I want to edit the code using recursion I'm always stuck at the first counting and can't figure out why. Any help is much appreciated..

public static int numberLength(int n) {
    if (n < 0) {
      n *= (-1);
    } else if (n == 0) {
      return 1;
    }

    int digits = 0;
    if (n > 0) {
      digits += 1;
      numberLength(n / 10);
    }
    return digits;

Upvotes: 0

Views: 8144

Answers (3)

jash
jash

Reputation: 13

A possible solution could look like this:

    public static int numberLength(int n){

    if(n < 0){
        return numberLength(-n);
    }

    else if(n == 0){
        return 0;
    }
    else{
        return 1 + numberLength(n/10);
    }
}

public static void main(String[] args){
    System.out.println(numberLength(-152555)); //returns 6
}

Upvotes: 0

Expired Data
Expired Data

Reputation: 678

In a recursive method you need to return some value based on reducing the size of the input value and combining this with your current count so far e.g.

public static int numberLength(int n){ 
   if(n < 10){ 
     return 1; 
   } 

   return 1 + (numberLength(n/10)); //This line combines the result
}

Upvotes: 2

ernest_k
ernest_k

Reputation: 45309

The problem is that you're discarding the result of numberLength(n / 10);

You probably meant to type:

int digits = 0;
if (n > 0) {
  return 1 + numberLength(n / 10);
}
return digits;

Upvotes: 0

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