Random numbers without duplicate

Question

public class startOrder {


public static void main(String[] args) {
int[] start = new int[45];

for(int i=0;i<start.length;i++)
{
    start[i] = (int)(Math.random()*45);
}

for(int i=0;i<start.length;i++)
{
    System.out.println(start[i]);
}
}

}

I want to make a simple program that just prints all random numbers between 1-45. The problem though is there's duplicate numbers among them, how can I make it to print numbers between 1-45 without any duplication. It's a program that generates a random start number.

Just add every already generated number to an array / list and check with `contains()` if that number has already been generated once - then skip it. Note: This may result in BAD performance if you like to generate random numbers from 1 to 40, while having bounds from 1 to 41. — dognose, Apr 06 '14 at 19:58

score 4 · Answer 1 · answered Apr 06 '14 at 20:05

4

Fill a list with the preferred numbers, then shuffle it, for example using the build in library function like below.

List<Integer> myNumbers = Arrays.asList(1,2,3,4, etc.);
Collections.shuffle(myNumbers);

Then simply iterate over the list.

Btw, this is called a random permutation.

answered Apr 06 '14 at 20:05

wvdz

16,251
4
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90

+1 This is the only answer that doesn't involve generating unnecessary random numbers for an unknown amount of time, losing efficiency. – leigero Apr 06 '14 at 20:17

score 0 · Answer 2 · answered Apr 06 '14 at 19:57

0

You could fill a linked list with 1-45, and repeatedly remove one randomly selected index from the list.

answered Apr 06 '14 at 19:57

earldouglas

13,265
5
41
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score 0 · Answer 3 · answered Apr 06 '14 at 19:57

0

Create an array or some collection, populate it with the numbers, then shuffle it using two randoms.

answered Apr 06 '14 at 19:57

Tony Hopkinson

20,172
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score 0 · Answer 4 · answered Apr 06 '14 at 20:02

0

Create a boolean array bArray of size 45 with all places set to false (which means that number is not picked yet).
Whenever you pick a number, let's say you picked 15, check if bArray[15] is false. If yes, pick that number and assign the bArray[15] to true (which means the number is now taken).

Repeat this till you have got all 45 numbers!

answered Apr 06 '14 at 20:02

Sarang Kane

36
3

Somewhat inefficient... – Tony Hopkinson Apr 06 '14 at 20:15

Azar · Answer 5 · 2014-04-06T20:14:13.597

0

You can use a HashSet!

public static void main(String[] args)
{
    Set<Integer> start = new HashSet<Integer>();

    while (start.size() <= 45)
    {
        start.add((int) (Math.random() * 45)+1);
    }

    for (int i: start)
    {
        System.out.println(i);
    }
}

A Set does not allow duplicates, so when you reach a size of 45, you have all unique numbers between 1 and 45! Also keep in mind that your random code generated a number between 0-44, which I fixed.

edited Apr 06 '14 at 20:14

answered Apr 06 '14 at 20:04

Azar

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Well it will work, eventually.... – Tony Hopkinson Apr 06 '14 at 20:13
@TonyHopkinson Yeah, not exactly efficient, but it sounds like a school assignment so I was trying to preserve the spirit of his code. – Azar Apr 06 '14 at 20:15

score 0 · Answer 6 · answered Apr 06 '14 at 20:05

0

Just use Set as such

Set<Integer> start = new HashSet<Integer>();

Set's are designed to eliminate duplicate values

answered Apr 06 '14 at 20:05

Maciej Cygan

5,351
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score 0 · Answer 7 · edited May 23 '17 at 10:33

You should look at the TreeSet and HashSet. These collections will not store any duplicates.

Random numbers without duplicate

7 Answers7