Java any digit Armstrong number

Issue

Following is my code for getting a three digit Armstrong number:

public class test {

    public static void main(String args[]) {

        for (int i = 100; i <= 999; i++) {
            int firstDigit = (i / 100);
            int secondDigit = (i % 100) / 10;
            int thirdDigit = (i % 10);
            if ((firstDigit * firstDigit * firstDigit)
                    + (secondDigit * secondDigit * secondDigit)
                    + (thirdDigit * thirdDigit * thirdDigit) == i) {

                System.out.println("this is a armstriong number - " + i);
            }
        }
    }
}

I am trying to get any digit Armstrong based on users input, but I am ending up writing too many loops and excess code.

Solution

The following code check if the scanned number (on console) is a Armstrong number. I've tested it, it works fine.

TEST IN CONSOLE

import java.util.Scanner;

class ArmstrongNumber
{
   public static void main(String args[])
   {
      int n, sum = 0, temp, remainder, digits = 0;

      Scanner in = new Scanner(System.in);
      System.out.println("Input a number to check if it is an Armstrong number");      
      n = in.nextInt();

      temp = n;

      // Count number of digits

      while (temp != 0) {
         digits++;
         temp = temp/10;
      }

      temp = n;

      while (temp != 0) {
         remainder = temp%10;
         sum = sum + power(remainder, digits);
         temp = temp/10;
      }

      if (n == sum)
         System.out.println(n + " is an Armstrong number.");
      else
         System.out.println(n + " is not an Armstrong number.");         
   }

   static int power(int n, int r) {
      int c, p = 1;

      for (c = 1; c <= r; c++) 
         p = p*n;

      return p;   
   }
}

Source here.

---

WITH A FUNCTION

If you need to use it as a function, please try this. The n param is the number you want to check. My function return true if it is an Armstrong number, else it returns false.

public boolean isArmstrongNumber(int n) {
    int sum = 0, temp = n, remainder, digits = 0;
    while (temp != 0) {
         digits++;
         temp = temp/10;
      }

      temp = n;

      while (temp != 0) {
         remainder = temp%10;
         sum = sum + power(remainder, digits);
         temp = temp/10;
      }
      if (n == sum) //Armstrong number
          return true;
       else //Not Armstrong number
          return false;
}

Answered By - Mistalis