CODEWITHSUNDEEP
javarandom
cw fei
cw fei

Reputation: 1564

Generating Unique random numbers effectively in Java

I want to generate unique random numbers from range 0 to 999,999.

In order to achieve that, I tried:

ArrayList<Integer> list = new ArrayList<Integer>();

        for (int i = 0; i < 999999; i++) { 
            list.add(new Integer(i)); // Add numbers from 0 - 999,999 into ArrayList
        }

        Collections.shuffle(list); // shuffle them

        for (int i = 0; i < 10; i++) {
            System.out.println(list.get(i)); // printed unique numbers
        }

The problem is the larger the number I want to generate, the longer the time it takes, for the above method, it took about 700ms.

But if I use Random() to generate them without filter duplicate numbers, it only takes 2ms

for(int i = 0; i<10; i++) {
  int digit = 0 + new Random().nextInt((999999 - 0) + 1); 
  System.out.println(digit);
}

Is there other way to generate unique random numbers in a more efficient manner?

Upvotes: 1

Views: 1790

Answers (3)

Kumaran
Kumaran

Reputation: 136

here is my code

its work perfect

private int count;

private boolean in = true;

public static final Random gen = new Random();

int[] result;

public int getCount() {
    return count;
}

public void setCount(int count) {
    this.count = count;
}

public void addUse() {

    result = new int[getCount() + 10];

    for (int i = 1; i < 10; i++) {

        String setup = ("" + i + i + i + i + i);

        result[getCount() + i] = Integer.valueOf(setup);

    }

}

public boolean chechArray(int number) {

    for (int i = 0; i < getCount() + 10; i++) {

        if (result[i] == number) {

            in = true;

            break;

        } else {

            in = false;

        }

    }
    return in;

}

public void printRandomNumbers() {

    Random gen = new Random();

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

        int get = gen.nextInt(100000 - 10000) + 10000;

        if (chechArray(get) == false) {

            result[i] = get;

        } else {

            i--;

        }

    }

}

public void viewArray() {

    printRandomNumbers();

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

        System.out.println((i + 1) + " Number is = " + result[i]);

    }

}

public static void main(String[] args) {

    RandomDemo3 rd2 = new RandomDemo3();
    rd2.setCount(20);
    rd2.addUse();
    rd2.viewArray();

}

Upvotes: 0

Paul Boddington
Paul Boddington

Reputation: 37655

There is no need to create a list of 1000000 numbers and shuffle them all if you only need 10. There is also no need to write new Integer(i) (you can just use i).

In Java 8 there is a very short way to do this:

int[] arr = ThreadLocalRandom.current().ints(0, 1000000).distinct().limit(10).toArray();
System.out.println(Arrays.toString(arr));

If you are using Java 7 or below, you could do this:

Random rand = new Random(); // Only do this in Java 6 or below. Now you should use ThreadLocalRandom.current().
int[] arr = new int[10];
Set<Integer> set = new HashSet<Integer>();
for (int index = 0, a; index < 10;)
    if (set.add(a = rand.nextInt(1000000)))
        arr[index++] = a;
System.out.println(Arrays.toString(arr));

Upvotes: 5

Justin D.
Justin D.

Reputation: 4976

You could create a set of random integers like this:

Set<Integer> set = new HashSet<Integer>();
Random rand = new Random();

while (set.size() < 10) {
    set.add(rand.nextInt((1000000)));
}

The idea is that the set data structure will remove duplicates.

Upvotes: 4

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