random unique number in java

Question

I'm trying to get a a list of random number and put it in queue without any repetition of the random number.

        int number = 40;

        for (int j = 0; j<number; j++)
        {
            int pick = random.nextInt(number);
            myQueue.add(Integer.toString(pick));
        }
        System.out.println("the queue size: "+myQueue.size());
        Iterator it = myQueue.iterator();
        while(it.hasNext()){
                String iteratorValue = (String)it.next();
                System.out.println("queue next value: "+iteratorValue);
        }

with the above code, I got some repetition of the random number

anyone know how to do it??

Are there any ranges on what random numbers you can generate? — templatetypedef, Jan 05 '13 at 18:42
possible duplicate of [Generating Unique Random Numbers in Java](http://stackoverflow.com/questions/8115722/generating-unique-random-numbers-in-java) — Kate Gregory, Jan 05 '13 at 20:25

arshajii · Accepted Answer · 2013-01-05T19:38:24.900

8

How about this:

List<String> list = new ArrayList<String>(number);

for (int i = 0; i < number; i++)
    list.add(Integer.toString(i));

Collections.shuffle(list);

myQueue.addAll(list);

"Adding unique random numbers" in some range is equivalent to adding all of the numbers in the range and then shuffling the result.

edited Jan 05 '13 at 19:38

answered Jan 05 '13 at 18:44

arshajii

127,459
24
238
287

anw, I got error in Collections.shuffle(myQueue). I have import the java.util.Collections...the error is "not suitable found for shuffle" – Newbie Newbae Jan 05 '13 at 19:04
`Collections.shuffle` only works for a `List`, not a `Queue`. Maybe shuffle a list first, then put that into the queue? – Louis Wasserman Jan 05 '13 at 19:05

Mike · Answer 2 · 2013-01-05T18:50:22.910

Create a Set and add you numbers to it as you generate them. The each time you generate a new number check if the Set already contains that value. Continue generating new numbers and checking the set until you find one that is not already there.

Something along these lines... (Note that Set.add(...) returns false if the value is already in the Set, so the do-while loop continues until a unique number is generated.)

   int number = 40;
   Set mySet = new HashSet();
   for (int j = 0; j<number; j++)
   {
       Integer pick;

       do{
           pick = random.nextInt(number);
       } while(!mySet.add(pick));
       myQueue.add(Integer.toString(pick));
   }
   System.out.println("the queue size: "+myQueue.size());
   Iterator it = myQueue.iterator();
   while(it.hasNext()){
           String iteratorValue = (String)it.next();
           System.out.println("queue next value: "+iteratorValue);
   }

Though as mentioned in by A. R. S., you don't seem to be looking for a random unique number, but rather a randomly shuffled list of all numbers between 0 and 40. If that is the case, use his/her solution as it is a better way to achieve that goal.

score 0 · Answer 3 · edited Jan 05 '13 at 19:05

0

If there is a small range of the random numbers, then you can simply generate list with available values and use Collections.shuffle on the list.

edited Jan 05 '13 at 19:05

Charles Menguy

40,830
17
95
117

answered Jan 05 '13 at 18:45

zibi

3,183
3
27
47

random unique number in java

3 Answers3

Linked

Related