A quicker method for finding out the number of digits in a number and print it?

Question

I have a question given by my computer teacher to find out the number of two digit number and three digit number and terminate the program (when user enter 0 )and print the total two digit ,three digit numbers and other digit number.

This is an example output in terminal window:

Enter your number: 23
Enter your number: 1
Enter your number: 412
Enter your number: -123
Enter your number: 32332
Enter your number: 12
Enter your number: -1
Enter your number: 0
Two digit numbers : 2
Three digit numbers : 2
Other digit numbers : 3

This is my program :

import java.io.*;
public class two_three_digits
{
    static DataInputStream dis = new DataInputStream(System.in);
    static void main()throws IOException
    {
        int two = 0 , three = 0 , oth = 0;
        while(true)
        {
            System.out.print("Enter your number: ");
            int n = Integer.parseInt(dis.readLine());
            n = (int)Math.abs(n);
            int d = 0,cn = n;
            if(n == 0)break;
            else if(n > 9 && n < 100)two++;
            else if(n > 99 && n <1000)three++;
            else oth++;
        }
        System.out.println("Two digit numbers : "+two);
        System.out.println("Three digit numbers : "+three);
        System.out.println("Other digit numbers : "+oth);
    }
}

I wished to know if there was any shorter way to do the same task ? Like using arrays , string or Bitswise operator , etc. Also I would rather like if you would tell the method of how to do it instead of simply the program as I am a learning student and wish to solve it out myself.

Edit: Sorry because i forgot to term one thing - by the word quicker i actually meant for speed (and more or less , size of program too)

Answer 1

An easy way to calculate the number of digits of a decimal number is to use the log base 10. This will give you a decimal value of what power to raise 10 to in order to get the input value.

For example, log10(10) would give 1, log10(100) would give 2, and values in between would give a value between 1 and 2. You could take the floor of this result, and add 1 to get the number of digits in the given number.

Answer 2

Since you're reading strings from the input, it would be less processing complex to just check that the string contains only digits and then take the string length as number of digits.

No parsing of the number actually entered needed.

String lineString = dis.readLine()
// check for valid input here (if needed)
int l = lineString.length();
if (lineString.charAt(0) == '-') // for negative numbers
    --l;
switch (l) {
    default:
        oth++;
        break;
    case 2:
        two++;
        break;
    case 3:
        three++;
        break;
}

Answer 3

Building on the idea of using log₁₀ to determine the number of digits, use an array to store the number of numbers with a certain number of digits.

public static void main(final String[] args) throws InterruptedException {
    int[] digits = new int[3];
    final Console c = System.console();
    while (true) {
        final int n = Math.abs(Integer.parseInt(c.readLine("Enter your number: ")));
        if (n == 0) {
            break;
        }
        digits[(int) Math.floor(Math.log10(n))]++;
    }
    System.out.println(Arrays.toString(digits));
}

Obviously if the user enters a number with more than three digits, the program will crash - so if you want to support that you could add bounds checks.

If by "quicker" you means "easier" then this solution certainly has fewer lines...