Add unique random numbers to an integer array

Question

I've created an ArrayList for integers which I would like to fill with 200 numbers. Each number can be within a range between 0 and 1023.

Therefore I've written this code:

Random rand = new Random();
ArrayList<Integer> values = new ArrayList<Integer>();

int START_AMOUNT = 200;

for(int i = 0; i < START_AMOUNT; 

  values.add(rand.nextInt(1024));      

}

As You might see, the for-loop will add 200 random numbers to the "values" ArrayList, from 0 to 1023. Now my problem is that I want the Array to have only unique numbers. How can I tell the Random class not to generate any numbers that already are existent in the ArrayList?

possible duplicate of Generating Unique Random Numbers in Java — Parimal Raj
– Parimal Raj, Commented Jul 29, 2014 at 9:27
you don't tell the Random class to do anything - that would make it non-random. — Alnitak
– Alnitak, Commented Jul 29, 2014 at 9:38

EpicPandaForce · Accepted Answer · 2014-07-29 11:19:09Z

4

What I'd do is creating an array of 1023 int composed by 1,2,3,...,1023. Then you shuffle it, and you take only the 200 first terms :

List<Integer> ints = new ArrayList<Integer>();
for(int i = 1; i <= 1023; i++)
{
    ints.add(i);
}

Collections.shuffle(ints);

EDIT as suggested by @Bohemian♦

List<Integer> result = ints.subList(0,200);

edited Jul 29, 2014 at 11:19

EpicPandaForce

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answered Jul 29, 2014 at 9:35

Kabulan0lak

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3 Comments

Bohemian Over a year ago

Lose everything after the first loop and use List.subList() instead: ints = ints.subList(0, START_AMOUNT);

Kabulan0lak Over a year ago

I didn't want to make mistake because I can't remember well Lists manipulation. So this works and he have the idea, but yes sublist seems pretty nice.

user2410644 Over a year ago

Pretty nice idea, I remember having it done similiar to this, thanks!

EpicPandaForce · Accepted Answer · 2014-07-29 09:52:27Z

2

A Set is a Collection that cannot contain duplicate elements. 
It models the mathematical set abstraction. 
The Set interface contains only methods inherited from Collection 
and adds the restriction that duplicate elements are prohibited.

And therefore,

public boolean add(E e)
  Adds the specified element to this set if it is not already present. 
  [...]
  If this set already contains the element, 
      the call leaves the set unchanged and returns false.

As such, what I'd do is use a Set, then add those to the list:

List<Integer> values = new ArrayList<Integer>();
Set<Integer> set = new HashSet<Integer>();
while(set.size() < 200) 
{
    set.add(rand.nextInt(1024)); 
}
values.addAll(set);

edited Jul 29, 2014 at 9:52

answered Jul 29, 2014 at 9:41

EpicPandaForce

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1 Comment

Athanor Over a year ago

I agree, you should use a set to avoid duplication.

JamesB · Accepted Answer · 2014-07-29 09:41:26Z

1

Use a Set:

Random rand = new Random();
Set<Integer> values = new HashSet<Integer>();
final int START_AMOUNT = 200;

while(values.size() < START_AMOUNT) { 
    values.add(rand.nextInt(1024));      
}
List<Integer> uniqueList = new ArrayList<Integer>(values);
System.out.println(uniqueList);

answered Jul 29, 2014 at 9:41

JamesB

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4 Comments

EpicPandaForce Over a year ago

I was faster by 7 seconds, heheh.

user2410644 Over a year ago

What's the difference between using a Set or another List?

EpicPandaForce Over a year ago

When you try to add an element into the Set using add() that is already contained by the Set, then the add() method will return false and not add the duplicate element.

JamesB Over a year ago

See Zhuinden answer above. A Set cannot contain duplicates.

ch1ll · Accepted Answer · 2014-07-29 09:40:24Z

0

You could also check if the ArrayList contains the given random number everytime you want to add one.

Random rand = new Random();
Integer r;
ArrayList<Integer> values = new ArrayList<Integer>();

int START_AMOUNT = 200;

for(int i = 0; i < START_AMOUNT; i++) {

  r = rand.nextInt(1024);
  If !values.contains(r) {
    values.add(r);
  } else {
   i--;
  }

}

Although i think Kabulan0lak's answer would be more performant if that is important.

answered Jul 29, 2014 at 9:40

ch1ll

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