Thread: Recursion Problem Need Help ASAP

I need help making a recursive program that computes the number of odd digits in a number. This is as far as I have gotten.

public class ComputeOddNumbers {
 
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
 
   System.out.println("Enter an integer:");
            Scanner input=new Scanner(System.in);
            int number= input.nextInt();
           computeoddnumbers(number); 
           System.out.print("There are "+odds+" odd numbers");
 
}
private static int odds;
    public static void computeoddnumbers(int number)
    {
        if(number<10)
        {
        increaseoddscount(number);
        }
        while(number>10)
        {
            increaseoddscount(number%10);
        }
 
 
 
 
    }
    public static int increaseoddscount(int number)
    {
 
 
       if(number==1||number==3||number==5||number==7|number==9)
       {
           odds++;
 
       }
 
 
        return  odds;
        }
 
    }

Heres the modified code

System.out.println("Enter an integer:");
            Scanner input=new Scanner(System.in);
            int number= input.nextInt();
           computeoddnumbers(number); 
           System.out.print("There are "+odds+" odd numbers");
 
}
private static int odds;
    public static void computeoddnumbers(int number)
    {
        if(number<10)
        {
        increaseoddscount(number);
        }
        else
        {
 
            {
       int numtemp =number%10;
       increaseoddscount(numtemp);
            }
 
        }
 
    }
    public static int increaseoddscount(int number)
    {
 
 
       if(number%2==1)
       {
           odds++;
       }
 
 
        return  odds;
        }
 
    }

Here's a nice little piece of pseudo-code that you can follow to create your code:

int countOdds(int number)
    If your number is less than 10:
        if it's odd, return 1
        else return 0
    else:
        if number is odd, return 1 + countOdds(number / 10)
        else return countOdds(number / 10)