android - generate random numbers with no repeat [duplicate]

Answer 1

Make a list of generated numbers, when your newly generated number is already in this list you make a new random number.

Random rng = new Random(); // Ideally just create one instance globally
List<Integer> generated = new ArrayList<Integer>();
for (int i = 0; i < numbersNeeded; i++)
{
    while(true)
    {
        Integer next = rng.nextInt(max) + 1;
        if (!generated.contains(next))
        {
            // Done for this iteration
            generated.add(next);
            break;
        }
    }
}

Answer 2

If there are only a few numbers, less than 100, I think I solution could be create a boolean array and once you get a number, set the position of the array to true. I don't think that it takes long time until all the numbers appear. Hope it helps!

Cheers!

Answer 3

I would suggest adding the numbers to an ArrayList<Integer> and then use Collections.shuffle() to randomize their order. Something like this:

ArrayList<Integer> number = new ArrayList<Integer>();
for (int i = 1; i <= 10; ++i) number.add(i);
Collections.shuffle(number);

Answer 4

My two cents

public Collection<Integer> getRandomSubset(int max,int count){
    if(count > max){
        throw new IllegalArgumentException();
    }
    ArrayList<Integer> list = new ArrayList<Integer>();
    for(int i =  0 ; i < count ;i++){
        list.add(i);
    }       
    Collections.shuffle(list);
    return list.subList(0, count);
}