can anyone please tell me how to generate random numbers with no repeat example
random (10) should(may) return 3,4,2,1,7,6,5,8,9,10 with no repeat
Thanks
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it will help you: stackoverflow.com/questions/4040001/…Thiru– Thiru2011-10-11 11:12:36 +00:00Commented Oct 11, 2011 at 11:12
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4 Answers
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);
2 Comments
Yogesh
liked it! a question, if i execute Collection.Shuffle twice or thrice will it provide different order?
Till Helge
I'm pretty sure that it will. I can't find a reason why it shouldn't. Easy enough to check though, eh? ;)
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;
}
}
}
1 Comment
Richard Le Mesurier
unfortunately this method doesn't scale very well
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);
}
1 Comment
Richard Le Mesurier
I like the
subList() enhancement in this solution - great for dealing out a hand of cards from a deckIf 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!
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