CODEWITHSUNDEEP
javarecursion
user1560265
user1560265

Reputation: 55

Better way to generate Armstrong Numbers recursively

Is there a more elegant way to write this recursively? Armstrong Numbers

PS: been out of school for 15 years this is not homework, just some code I am trying to convert from iterative to recursively.

import java.util.Scanner;

public class RecArmstrong {
public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);

    System.out.print("Enter a number: ");
    int number = keyboard.nextInt();

    //Error checking
    while(number < 0 || number > 100000){
        System.out.print("Enter a number: ");
        number = keyboard.nextInt();
    }

    if(arm(number) == number)
        System.out.println(number + " is an armstrong number");
    else
        System.out.println(number + " is not an armstrong number.");

}

public static long arm(long n){

    long temp, sum, digits = 0;
    long remainder;
    temp = n;
    sum = 0;

    if (temp == 0)
        return 0; //base case
    else{
        while (temp != 0){
        digits++; //number of digits for exponent
        temp = temp / 10;
    }
    temp = n; //set temp back to original number
    while (temp != 0){
        remainder = temp % 10;
        sum += Math.pow(remainder, digits);
        temp = temp / 10;
    }
        return sum + arm(temp);
    }
}
}

Upvotes: 2

Views: 3185

Answers (2)

Lalit Behera
Lalit Behera

Reputation: 540

If you want to do it using java stream, you can do it like this.

public boolean isArmstrongNumber(String number) {
    int exponent = number.length();
    if (Integer.parseInt(number) == number.chars()
            .map(n -> n - '0')
            .map(n ->(int) Math.pow(Integer.parseInt("" + n),exponent))
            .sum())
        return true;
    else return false;
}

Upvotes: 0

Legionair
Legionair

Reputation: 423

Maybe im mistaken (its 1AM here..), but as far as i can tell your implementation is actually not a recursion at all. With

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

you do the whole calculation iteratively, until tmp is 0. Therefore in the next line

return sum + arm(temp);

arm(temp) will always return 0!

I hacked together a quick recursion of my own, which works digit by digit, starting from the last. The function overload is necessary because every recursive call needs the total length of the original number.

public static long arm(long n){
   return arm(n, Integer.toString(n).length());
}
public static long arm(long n, int num_digits){

   if(n==0) //recursion finished
      return;
   // n%10 gives last digit
   return java.lang.Math.pow(n%10,num_digits) + arm(n/10, num_digits);
}

I hope you like it ;-)

Upvotes: 4

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