Concat String in a for loop

Question

I'm trying to create a program that takes each number typed in by the user and sort them out as even, odd and the number zero values.

The result should look like something like this:

User Input: 14005

Output:
Even Numbers: 4
Odd Numbers: 1, 5
Zero's: 0, 0

This is the code I've written, I thought of using string concatination in order to add a new value each time the loop checks for the next character, don't know whether I'm thinking right or not though, would appriciate if someone could tell me where I'm thinking in the wrong way.

package com.craydesign;

import javax.swing.JOptionPane;

public class Main {

    public static void main(String[] args) {
        String number = JOptionPane.showInputDialog("Please enter a number: ");
        String evenNumbers = "";
        String oddNumbers = "";
        String numberZero = "";

        for(int i = 0; i < number.length(); i++) {
            if(number.charAt(i) % 2 == 0) {
                evenNumbers.concat(Integer.toString(i) + ", ");
            } else if(number.charAt(i) % 2 != 0) {
                oddNumbers.concat(Integer.toString(i) + ", ");
            } else if (number.charAt(i) == 0){
                numberZero.concat(Integer.toString(i) + ", ");
            }
        }

        JOptionPane.showMessageDialog(null, "Even numbers: " + evenNumbers + "\n" + "Odd numbers: " + oddNumbers + "\n" + "Zero's: " + numberZero);

    }

}

If you told the exact problem/output of your current code, it would be easier to help. — 4386427
– 4386427, Commented Sep 19, 2015 at 11:13
String is immutable. This means that none of its methods change it. They simply return a new value. So String.concat() doesn't add anything to the string, it just returns a concatenated string. So you shouldn't use that, you should use StringBuilder instead. — RealSkeptic
– RealSkeptic, Commented Sep 19, 2015 at 11:13
@kemkoi see the @TAsk answer. You are dividing the ASCII value. Instead of String, use StringBuilder. — SatyaTNV
– SatyaTNV, Commented Sep 19, 2015 at 11:20
use ArrayList<String> instead of String. we dont need to concate. We can easily add the values using add() function — balaraman
– balaraman, Commented Sep 19, 2015 at 11:21

Rustam · Accepted Answer · 2015-09-19 11:37:04Z

1

use Character.getNumericValue() instead of charAt(i)

public static void main(String[] args) throws IOException
    {

          String number = JOptionPane.showInputDialog("Please enter a number: ");
          StringBuffer evenNumbers = new StringBuffer();
          StringBuffer oddNumbers =new StringBuffer();
          StringBuffer numberZero =new StringBuffer();

        for(int i = 0; i < number.length(); i++) {
          int value=Character.getNumericValue(number.charAt(i));
          if(value!=0 && value % 2 == 0) {
              evenNumbers.append(value).append(',');
          } else if(value % 2 != 0) {
              oddNumbers.append(value).append(',');
          } else if (value == 0){
              numberZero.append(value).append(',');
          }
        }

          JOptionPane.showMessageDialog(null, "Even numbers: " + evenNumbers + "\n" + "Odd numbers: " + oddNumbers + "\n" + "Zero's: " + numberZero);


      }

EDIT:(displaying numbers in sorted order)

     String evenNo[]=evenNumbers.toString().split(",");
      String oddNo[]=oddNumbers.toString().split(",");

      Arrays.sort(evenNo);
      Arrays.sort(oddNo);

      JOptionPane.showMessageDialog(null, "Even numbers: " +  Arrays.toString(evenNo) + "\n" + "Odd numbers: " + Arrays.toString(oddNo) + "\n" + "Zero's: " + Arrays.toString(numberZero.toString().substring(0, 

numberZero.length()-1).split(",")));

edited Sep 19, 2015 at 11:37

answered Sep 19, 2015 at 11:21

Rustam

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5 Comments

Jacek Cz Over a year ago

Change evenNumbers.append(value+","); to evenNumbers.append(value).append(","); - in sense of SB

kemkoi Over a year ago

This solved it, apparently I could've used either StringBuffer or StringBuilder. Didn't know String's were immutable, thanks for the answer!

Rustam Over a year ago

@kemkoi If you want to display numbers in sorted order you can try my Edit.

Rustam Over a year ago

@JacekCz yes can be done but that would be unnecessary append() method call. yet I updated.

Jacek Cz Over a year ago

better is one call (highly optimized in JVN ) that construction of 2-3 objects

Aman Agrawal · Accepted Answer · 2015-09-19 11:21:04Z

0

You are using String for output and appending the output in the string.

Its a bad idea. as String class is immutable so if you do change anything in String it will create a new Object in memory.

So your solution will take extra memory.

Accroding to me you can solve this problem in two ways

Use StringBuffer class instead of String class for appending
Use ArrayList to Store your result. and iterate over the arraylist and show the output as you want.

Thanks, Aman

answered Sep 19, 2015 at 11:21

Aman Agrawal

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