逐行打印数字位数的递归方法

Question

I'm supposed to use a recursive method to print out the digits of a number vertically.

For example, if I were to key in 13749, the output would be:

1
3
7
4
9

How should I approach this question?? It also states that I should use the if/else method to check for the base case.. I just started learning java and I'm not really good at it :(

import java.util.Scanner;

public class test2 {
  public static void main (String [] args){
    Scanner sc = new Scanner(System.in);
    System.out.print("Enter a positive integer: ");
    int n = sc.nextInt();
    System.out.println();
    System.out.println(numbers(n));  

  } 

public static int numbers(int n){
  int sum;
  if (n == 0) {
    sum = 1;
    } else {

     System.out.println(n%10);
     sum = numbers(n / 10) + (n % 10);


    }
  return sum;
  }      
}

Answer 1

I was asked this question today in an interview!

 public class Sandbox {
   static void prints(int d) {
        int rem = d % 10;
        if (d == 0) {
            return;
        } else {
            prints(d / 10);
        }
        System.out.println(rem);
    }

    public static void main(String[] args) {
        prints(13749);
    }
}

Output:

1
3
7
4
9

Answer 2

You asked how to approach this, so I'll give you a tip: it would be a lot easier to build up the stack and then start printing output. It also doesn't involve manipulating strings, which is a big plus in my book. The order of operations would be:

1. Check for base case and return if it is
2. Recursive call
3. Print

This way when you get to the base case, you'll start printing from the tail to the head of the calls:

recursive call 1
    recursive call 2
        recursive call 3
            .... reached base case
        print 3
    print 2
print 1

This way you can simply print number % 10 and make the recursive call with number / 10, the base case would be when number is 0.

Answer 3

class PrintDigits {

   public static void main(String[] args) {
     String originalNumber, reverse = "";

     // Creating an Scanner object
     Scanner in = new Scanner(System.in);

     System.out.println("Enter a number:");
     // Reading an input 
     originalNumber = in.nextLine();

     // Calculating a length
     int length = originalNumber.length();

     // Reverse a given number
     for ( int i = length - 1 ; i >= 0 ; i-- )
        reverse = reverse + originalNumber.charAt(i);
     //System.out.println("Reverse number: "+reverse);
     digits(Integer.parseInt(reverse));
   }

   /* digits of num */
   public static void digits(int number) {
       if (number == 0)
          System.out.println("");
       else {
          int mode=10;
          System.out.println(+number%mode);
          digits(number/mode);
       }
   }
}

Answer 4

Here is my code in C++

Just modify it for Java. You need to show the number after you call the function that way you show the last one first... as per the answer from s.ts

void recursive(int n) {
    if (n < 10)
        cout << n << endl;
    else
    {
        recursive(n / 10);
        cout << n % 10 << endl;
    }
}

Answer 5

If number consists of more than one digit print ( n / 10 )

print ( n % 10 )

If you want them printed in the other order

print ( n % 10 )

If number consists of more than one digit print ( n / 10 )

Answer 6

try this

public class Digits {

    public static void main(String[] args) {
        printDigits(13749);
    }

    private static void printDigits(Integer number) {
        int[] m = new int[number.toString().toCharArray().length];

        digits(number, 0, m, 0);

        for (int i= m.length - 1; i>=0; i--) {
            System.out.println(m[i]);
        }
    }

    public static int digits(int number, int reversenumber, int[] m, int i) {

        if (number <= 0) {
            return reversenumber;
        }

        m[i] = (number % 10);
        reversenumber = reversenumber * 10 + (number % 10);
        return digits(number/10, reversenumber, m, ++i);
    }

}

Answer 7

Python example

def printNoVertically(number):
    if number <  9:
      print(number)
      return
    else:
        printNoVertically(number/10)
        print(number % 10)