how to search for the same element in an ArrayList?

Question

This Code creates 50 random-numbers between 1 and 100 and adds it in an ArrayList. Now I want to search the ArrayList for the same numbers, than remove them and get a new number. In the end, the list should only contain 50 unique numbers between 1 and 100.

The Problem is: I don't know, how to search the same ArrayList for the same number, remove it and get a new one. Can someone please help me?

import java.util.Random;
import java.util.ArrayList;

class RandomPrim {

    public static void main(String[] args) {

        Random nr = new Random();
        int number;
        ArrayList<Integer> liste = new ArrayList<Integer>();

        // get 50 random numbers between 1 and 100
        for(int counter = 1; counter <= 50; counter++) {
            number = 1+nr.nextInt(100);
            liste.add(number);

            // System.out.println(liste.toString());
        }

        for (int ausgabe : liste) {
            System.out.print(ausgabe+ ", ");
        }
    }
}

As @gevorg said you can use a HashSet and add random numbers to it. You can stop when its length is 50. — xro7
– xro7, Commented Jun 30, 2016 at 13:00
Another approach could be to create a list of 100 numbers and randomly take one of the elements out of that list, i.e. you'd get the next index by calling nextInt( availableNumbers.size() ) (ofc you should remove any number you took from that list). — Thomas
– Thomas, Commented Jun 30, 2016 at 13:02

Mr.7 · Accepted Answer · 2016-06-30 13:05:54Z

3

Random nr = new Random();
int number;
ArrayList<Integer> liste = new ArrayList<Integer>();

// get 50 random numbers between 1 and 100
for(int counter = 1; counter <= 50; ) {
    number = 1+nr.nextInt(100);
    if(!(liste.contains(number))) {
        liste.add(number);
        counter++;
    }
}

for (int ausgabe : liste) {
    System.out.print(ausgabe+ ", ");
}

hope this helps.

answered Jun 30, 2016 at 13:05

Mr.7

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Comments

gevorg · Accepted Answer · 2016-06-30 13:01:03Z

2

Better use HashSet instead, to avoid duplications:

Random nr = new Random();
int number;

Set<Integer> randomSet = new HashSet<>();

// get 50 random numbers between 1 and 100
while(randomSet.size() < 50) {
    number = 1 + nr.nextInt(100);
    randomSet.add(number);
}

for (int ausgabe : randomSet) {
    System.out.print(ausgabe + ", ");
}

answered Jun 30, 2016 at 13:01

gevorg

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1 Comment

Freeze Over a year ago

unfortunally I have to use an ArrayList or LinkedList

Thomas · Accepted Answer · 2016-06-30 13:10:32Z

As already has been suggested using a Set or, if insertion order is important, a LinkedHashSet and keep generating random numbers until the set has a size of 50.

The problem with that, however, might be that as the set fills up you could get more and more duplicates thus requiring a lot of retries.

So an alternative could be to use a list of 100 numbers and then randomly take one out:

List<Integer> availableNumbers = new ArrayList<>( 100 );
for( int i = 1; i <= 100; i++ ) {
  availableNumbers.add( i );
}

Random r = new Random();
List<Integer> randomizedList = new ArrayList<>( 50 );
for( int i = 0; i < 50; i++ ) {
  int randomIndex = r.nextInt( availableNumbers.size() );
  randomizedList.add( availableNumbers.remove( randomIndex ) );
}

Note that using an ArrayList has the drawback that if you take out a number all numbers following it would have to be shifted to the left. On the other hand using a LinkedList would require an iteration to reach the index of the element that should be removed.

austin glover · Accepted Answer · 2016-06-30 13:05:07Z

1

You could use a linq query for this:

liste.Distinct().ToArray();

This should at least get you headed in the right direction.

answered Jun 30, 2016 at 13:05

austin glover

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Comments

MichaelQ · Accepted Answer · 2016-06-30 13:58:26Z

1

If you don't want to try any of the other solutions, you can just get unique numbers from the beginning while still using an ArrayList like so..

Random nr = new Random();
int number;
ArrayList<Integer> liste = new ArrayList<Integer>();

// get 50 random numbers between 1 and 100
for(int counter = 1; counter <= 50; ) {
    number = 1+nr.nextInt(100);
    if(!(liste.contains(number))) {
        liste.add(number);
        counter++;
    }
    else {
        counter = counter - 1;
    }
}

for (int ausgabe : liste) {
    System.out.print(ausgabe+ ", ");
}

edited Jun 30, 2016 at 13:58

answered Jun 30, 2016 at 13:11

MichaelQ

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1 Comment

Thomas Over a year ago

This will use recursion - how so? I don't see any recursive call. Also counter = counter - 1; might result in an endless loop.

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