Discarding array item once used

Question

How can I make sure that a random number is only generated once?

For example:

for(int i = 0; i<len; i++)
    {
        while (rands.contains(rand = r.nextInt(len-2)+1));
        rands.add(rand);
        System.out.print ("Rand ___ " + rand + "___");
    }

How can I make sure that if the number 2 is generated from rand, it wont be generated again?

I apologise if i am not making myself clear enough. Please comment if you require any more information.

Thanks.

What are you actually trying to do? If you need the numbers 0-2 generated in a random order you might want to just use Collections.shuffle(List). — Mike Deck, Jul 30 '12 at 19:39
You can probably find your answer [here](http://stackoverflow.com/questions/196017/unique-random-numbers-in-o1), as the same question is asked. — Viktor, Jul 30 '12 at 19:39
Of course, once you start excluding previously used numbers it's no longer *random* — Stephen P, Jul 30 '12 at 21:24

score 1 · Answer 1 · answered Jul 30 '12 at 19:37

1

You can't really constrain that when using Random API. May be you need to keep a list of already generated numbers and return the number if it is not already generated.

answered Jul 30 '12 at 19:37

kosa

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is there any other way to do it? without Random API? – faraday Jul 30 '12 at 19:38
could i create a while loop that carries on running until its not equal to the numbers previously generated? – faraday Jul 30 '12 at 19:39
Another way may be, have a list (or) array of numbers and get number from that list (or) array. Keep another array for already visited index. If list, may be remove already visited number. – kosa Jul 30 '12 at 19:39
1

*Don't* use a list or array to store the past numbers; that would be the slowest way to do it. Use a `HashSet` or something similar. – Ernest Friedman-Hill Jul 30 '12 at 19:41
HashSet is collection/datastructure implementation. Read this link for more info. http://docs.oracle.com/javase/6/docs/api/java/util/HashSet.html – kosa Jul 30 '12 at 19:45
ok then i think this might work. Ill try right now and keep you all posted! – faraday Jul 30 '12 at 19:48
pleas help i dont know why its doing this – faraday Jul 30 '12 at 19:59

score 1 · Accepted Answer · answered Jul 30 '12 at 19:39

1

You'd have to store all the numbers you've generated, and check the new ones. Use a HashSet for performance' sake:

HashSet<Integer> rands = new HashSet<Integer>();
for(int i = 0; i<3; i++)
{
   int rand;
   while (rands.contains(rand = r.nextInt(3)))
       ;
   rands.add(rand);
}

answered Jul 30 '12 at 19:39

Ernest Friedman-Hill

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`java.util.HashSet`. It's a class that can store and retrieve a set of unique objects in (amortized) constant time. It's a great way to keep track of whether an object has previously come up in a sequence. – Ernest Friedman-Hill Jul 30 '12 at 19:46
can you reset it? like an array – faraday Jul 30 '12 at 19:48
please help, im not sure why its doing this – faraday Jul 30 '12 at 19:59
In my loop condition, I'm setting `rand` to hold each successive random number, but you don't seem to be doing that; also, I'm testing that the set contains `rand`, but you're testing for some other value altogether. – Ernest Friedman-Hill Jul 30 '12 at 20:40
Is there a problem now? I tried your code with a proper set of declarations preceding it, and it compiles and correctly generates a set of distinct random integers. – Ernest Friedman-Hill Jul 30 '12 at 20:47
the program hangs for some reason – faraday Jul 30 '12 at 20:53

score 0 · Answer 3 · answered Jul 30 '12 at 19:40

0

I'm assuming that you want to create a list of numbers from 0-2 in random order (If I'm wrong, let me know)

What you really want is to create a list of those numbers

ArrayList<Integer> randomNums = ArrayList<Integer>() {{ add(0); add(1); add(2); }}

Then use Collections to shuffle that.

Collections.shuffle(randomNums)

answered Jul 30 '12 at 19:40

SomeKittens

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well the number of numbers (a bit tongue twistery) can change – faraday Jul 30 '12 at 19:43

score 0 · Answer 4 · answered Jul 30 '12 at 19:44

You need to use a shuffling algorithm for this. Start with 1. Create an array of Integers from 1 to 1000 (arbitrary Size). Shuffle the array well. keep fetching from the shuffled array till you run out of them. When you are through the 1000, repopulate the array with 1000-2000, shuffle, and use. Repeat the process as many times as you need.

score 0 · Answer 5 · answered Jul 30 '12 at 19:46

If you just need a stream of non-repeating random numbers in some range you could do something like this:

import java.util.ArrayList;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;

public class RandomNumberStream {
    private Iterator<Integer> numbers;

    public RandomNumberStream(int range) {
        List<Integer> numbersList = new ArrayList<>(range);
        for(int i = 0; i < range; i++) {
            numbersList.add(i);
        }
        Collections.shuffle(numbersList);
        numbers = numbersList.iterator();
    }

    public int nextNumber() {
        if(!numbers.hasNext()) {
            throw new IllegalStateException("No more numbers...");  //you might want to handle this differently
        }
        return numbers.next();
    }

    public static void main(String[] args) {
        RandomNumberStream rand = new RandomNumberStream(5);

        for(int i = 0; i < 5; i++) {
            System.out.println(rand.nextNumber());
        }
    }
}

Discarding array item once used

5 Answers5