Question

Please have a look at the following code. It is my attempt to manage the given numbers in ascending order.

import java.io.*;
import java.util.*;
import java.util.ArrayList;

public class TurboSort
{
    public static void main(String[]args)
    {

        List<Integer> numbers = new ArrayList();        
        Scanner scan = new Scanner(System.in);

        while(scan.hasNextInt())
        {
            numbers.add(scan.nextInt());
        }


        Collections.sort(numbers);

        System.out.println(numbers);
    }
}

insert the input as 2,1,6,7,3

Hit enter.

Now, the scanner hasn’t exited from the while loop because it is not giving any output. What am I doing here wrong? Even if you manage to get it, the output is surrounded by brackets like ” [1] [2] [3] “. Why is that? Is that is because I didn’t call ‘Integer.parseInt()’ ?. Please help me with those 2 questions.

Thanks.

Answer 1

This loop will never exit (as long as you enter integers) as there is no break condition

while(scan.hasNextInt()){
  numbers.add(scan.nextInt());
}

If you want your loop to stop, say for example you need to acquire only 5 integers then you could do this:

while(scan.hasNextInt()){
  numbers.add(scan.nextInt());
  if(numbers.size() == 5) break;
}