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Home/ Questions/Q 7716291
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T02:35:57+00:00 2026-06-01T02:35:57+00:00

I am trying to split a user inputed set, such as { 1 2

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I am trying to split a user inputed set, such as { 1 2 3 4 }, as a string, into an array list so that when I print out the array list it will read {1, 2, 3, 5}. Here is my code so far. I am not really sure how the Scanner.next() method works, but I am attempting to use it. This is for a small portion of my program. Actually it’s like the beginning.

import java.util.Scanner;
import java.util.ArrayList;
public class practice {
   public static void main (String[]args){
    Scanner stdIn = new Scanner (System.in);
    String set;
    String set2 = "";

System.out.print("Enter set:");
set = stdIn.nextLine();
if(set.charAt(0) == '{'){
    for(int i =1; i<set.length(); i++){
        set2 += set.charAt(i);
    }
}
else if(set.charAt(1) == '{'){
    for(int i = 2; i <set.length();i++){
        set2 += set.charAt(i);
    }
}
System.out.print(set2);


ArrayList<Integer> array = new ArrayList<Integer>();

while(stdIn.next() != "}"){
    set2 = stdIn.next();
    array.add(Integer.parseInt(set2));
}
System.out.print(array);

  }
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T02:35:59+00:00Added an answer on June 1, 2026 at 2:35 am

    You should use the String.split(String) and String.replaceAll(String, String) methods in String in order to do a lot of what your code currently does.

    You can get the user input as you are currently doing it, with stdIn.nextLine();, but afterward, you can do the string processing in an easier way.

    First, you should remove the unnecessary characters, and end up with just a sequence of numbers separated by spaces.

    You can do this by simply calling the replaceAll method, and provide the regular expression, which is really simple in this case.

    Then you can call the String.split method to find each element in the set.

    String in = stdIn.nextLine();
    // Need to escape the characters because
    // these characters mean something special in regular expressions
    String filtered = userInput.replaceAll("(\\[\\{\\}\\],"," ");
    String[] numbers = filtered.split("\\s+");
    ArrayList <Integer> myNumbers = new ArrayList <Integer> ( numbers.length );
    for ( String number : numbers )
        myNumbers.add ( Integer.parseInt ( number ) );
    

    UPDATE

    If you need to use a scanner, you can use a scanner for the String that has been filtered

    Scanner s = new Scanner ( filtered );
    ArrayList <Integer> myNumbers = new ArrayList < Integer > ();
    while ( s.hasNext() )
    {
       myNumbers.add(s.nextInt());
    }
    
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