如何在ArrayList中搜索相同的元素？

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+ ", ");
        }
    }
}

Answer 1

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.

Answer 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 + ", ");
}

Answer 3

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.

Answer 4

You could use a linq query for this:

liste.Distinct().ToArray();

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

Answer 5

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+ ", ");
}