Select a random int that doesn't exist

Question

I am doing something where I have to choose a couple random numbers, without choosing the same number again. I have tried many ways, but they will not work. I checked if the random number exists in my int[], and reset the int to another random, but what is that other random also exists, I tried fixing that but I ran into problems.

Here's my current code:

p.sendMessage("debug over max");
        Random r = new Random();
        for (int i=0;i<max + 1;i++) {
            int ran = r.nextInt(arenaAmount);
            if (ran == 0) ran = 1;
            arenas[i] = ran;
        }

Thats what I have so far, so how can I make sure it doesn't have the same number. If there is another thread already please link me to it.

Thanks, Joey.

can't you pick the previous random number again? or any of the previous numbers? — Gabriel Espinel, Feb 06 '15 at 19:24
Or set a Random range to exclude (either above or below) the previously chosen random integer? — Phoenix, Feb 06 '15 at 19:25
[This](http://codereview.stackexchange.com/questions/61338/generate-random-numbers-without-repetitions) might be useful. — Emil Laine, Feb 06 '15 at 19:27
[This](http://stackoverflow.com/questions/8115722/generating-unique-random-numbers-in-java) might also help. — Giovanni Botta, Feb 06 '15 at 19:37

score 1 · Answer 1 · edited May 23 '17 at 12:09

1

A simple solution would be to add the already generated numbers to a Set and generate random numbers until you hit one that isn't already in that Set.

But that's probably not a very good solution, check the accepted answer here for a thorough explanation.

As mentioned by Giovanni Botta in the comments, here's another simple solution that's probably better than the Set based one.

edited May 23 '17 at 12:09

Community

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1

answered Feb 06 '15 at 19:34

Emil Laine

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@ColinD True, added it to my answer. – Emil Laine Feb 06 '15 at 19:40

score 0 · Answer 2 · answered Feb 06 '15 at 19:34

0

make a arenas a Set

Random r = new Random(); 
int ran = r.nextInt(); 
while( ! arenas.add(ran) ) {
   ran = r.nextInt(); 
}

add will fail on an attempted re-entry of a value.

answered Feb 06 '15 at 19:34

corn3lius

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I tried that, still didn't work. – Beta Nyan Feb 06 '15 at 19:54

Anderson Vieira · Answer 3 · 2015-04-22T11:32:02.237

You could create a list of integers from 1 to maxValue, shuffle it and get the numElements first elements:

List<Integer> shuffledList(int maxValue, int numElements) {
    if (numElements >= maxValue) {
        throw new IllegalArgumentException("The number of elements in the list must be less than maxValue.");
    }
    List<Integer> numbers = range(1, maxValue);
    Collections.shuffle(numbers);
    return numbers.subList(0, numElements);
}

List<Integer> range(int from, int to) {
    List<Integer> numbers = new ArrayList<>(to - from);
    for (int i = from; i < to; i++) {
        numbers.add(i);
    }
    return numbers;
}

This way, you are sure to get different numbers without the Set overhead. Making a call like shuffledList(10, 5) would, for example, return a list like [8, 7, 5, 1, 2], with 5 elements, where the smallest possible element is 1 and the greatest possible element is 9.

Also, if you are using Java 8 you can discard the range function and do this instead:

List<Integer> numbers = IntStream.range(1, maxValue)
                                 .boxed()
                                 .collect(Collectors.toList());

Select a random int that doesn't exist

3 Answers3

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