How to generate 20 unique random numbers with order using Java?

Question

am new to programming and I just started doing a program where I want to generate and sort 20 unique random numbers, how ever I was only able to generate numbers which were not unique nor sorted using this script

import java.util.Random;

class FTW {
    public static void main (String[]args){
        Random I  = new Random();
        int number; 

        for(int counter=1; counter<=20;counter++){
            number = I.nextInt(20);
            System.out.println(number + " ");
        }
    }
}

Can any one help me edit this or can give me a better one that does the job and explain it to me if possible :)

http://stackoverflow.com/questions/8115722/generating-unique-random-numbers-in-java — SatyaTNV, Jul 27 '15 at 14:11

score 2 · Accepted Answer · answered Jul 27 '15 at 14:18

One way is to add the numbers to an ArrayList and check if it contains the next random number in a while-loop. (These are 20 unique random numbers from 0-100)

public class FTW {
    public static void main (String[]args){
        Random I  = new Random();
        List<Integer> list = new ArrayList<Integer>();
        int number; 

        for(int counter=1; counter<=20;counter++){
            number = I.nextInt(100);
            while(list.contains(number)) {
                number = I.nextInt(100);
            }
            list.add(number);
        }
        Collections.sort(list); //Sorts the list
        System.out.println(list);
    }
}

Huang Chen · Answer 2 · 2015-07-27T14:42:02.250

0

Using a HashSet will preserve the uniqueness

This will provide a random set of 20 numbers from 1-50

  public static void main(final String[] args){
        final Random random = new Random();
        final Set<Integer> intSet = new HashSet<>();
        while (intSet.size() < 20) {
            intSet.add(random.nextInt(50) + 1);
        }
        final int[] numbers = new int[intSet.size()];
        final Iterator<Integer> iter = intSet.iterator();
        for (int i = 0; iter.hasNext(); ++i) {
            numbers[i] = iter.next();
        }
        System.out.println(Arrays.toString(numbers));
    }

edited Jul 27 '15 at 14:42

answered Jul 27 '15 at 14:14

Huang Chen

1,177
9
24

@ctwomey1 sorry didn't mean order, meant uniqueness, will change now – Huang Chen Jul 27 '15 at 14:41

Razib · Answer 3 · 2015-07-27T14:52:00.650

You may follow these steps:
1. maintain an ArrayList to store the generated unique random numbers (uniqueSortedRandoms).
2. While inserting a new number newNumber into this ArrayList, check if the array uniqueSortedRandoms contains the newNumber and whether the newNumber is greater than or equals to the previous one. I as assuming the array is sorted ascending order. See the following code -

import java.util.Random;

class FTW {
    public static void main (String[]args){
        Random I  = new Random();
        int newNumber;
        List<Integer> uniqueSortedRandoms = new ArrayList<Integer>();


        for(int counter=1; counter<=20;){

            int previousNumber = -1 // initially set to -1
                                    // because nextInt() can be 
                                    //range from 0 (inclusive) to 20 exclusive
            newNumber = I.nextInt(20);

            if(newNumber>previousNumber && !unqueSortedRandoms.contains(newNumber)){

               uniqueSortedRandoms.add(newNumber);
               previousNumber = newNumber;
               counter++;  
            }

            System.out.println(number + " ");
        }
    }
}

Now the ArrayList - uniqueSortedRandoms contains all the unique random numbers you need in an ascending order.

Note:
1. If you use random.nextInt(20) then it will generate a random number from 0 inclusive to 20 exclusive. And you need 20 random numbers in a sorted order. So the array List actually contains 20 numbers from 0 to 19. In this case you can just generate the arrayList with numbers from 0 to 19.Or if you need 20 random numbers in sorted order (not in the range 0 to 19) like [3, 4, 9, 10, ......] then you may use a very large int as a parameter of nextInt(int n) -

newNumber = I.nextInt(100);

Now the each newNumber will be in the range - 0<=newNumber<100. So your array will contains 20 unique random number in a ascending order.

2. The counter is incremented inside the if-block (when the newly generated random number is inserted int the ArrayList) so that the loop continues until we get the 20th random number.

thanks a lot guys it worked perfectly :) :) and about the name FTW is because i tried too many times and had soo many class names i ended up with even weirder names than this xD — Tim, Jul 27 '15 at 15:07
@Tim, in this case you may consider to upvote or accept the answer :) — Razib, Jul 28 '15 at 03:26

score 0 · Answer 4 · answered Jul 27 '15 at 14:58

There is way how to generate unique, sorted numbers without sorting. You can create array of boolean and which number you choose, that index is true.

public static void main(String[] args) {
        int maxNumber = 150;
        int totalCount = 20;        
        Random random = new Random();

        boolean[] generatedNumbers = new boolean[maxNumber];
        int generatedCount = 0;

        while (generatedCount < totalCount){
            int newNumber = random.nextInt(maxNumber);
            if (generatedNumbers[newNumber] == false){
                generatedNumbers[newNumber] = true;
                generatedCount++;
            }
        }

        int[] sortedUniqueArray = new int[totalCount];

        int selectedNumbers = 0;
        for (int i = 0; i < generatedNumbers.length; i++) {
            if (generatedNumbers[i] == true){
                sortedUniqueArray[selectedNumbers] = i;
                selectedNumbers++;                
            }
        }

        System.out.println(Arrays.toString(sortedUniqueArray));
    }

Output for this sample is :

[6, 19, 33, 47, 51, 53, 71, 75, 82, 86, 89, 92, 105, 108, 121, 125, 126, 137, 140, 147]

How to generate 20 unique random numbers with order using Java?

4 Answers4