Generating Unique Random Numbers in Java

Question

I\'m trying to get random numbers between 0 and 100. But I want them to be unique, not repeated in a sequence. For example if I got 5 numbers, they should be 82,12,53,64,32

Answer 1

Choose n unique random numbers from 0 to m-1.

int[] uniqueRand(int n, int m){
    Random rand = new Random();
    int[] r = new int[n];
    int[] result = new int[n];
    for(int i = 0; i < n; i++){
        r[i] = rand.nextInt(m-i);
        result[i] = r[i];
        for(int j = i-1; j >= 0; j--){
            if(result[i] >= r[j])
                result[i]++;
        }
    }
    return result;
}

Imagine a list containing numbers from 0 to m-1. To choose the first number, we simply use rand.nextInt(m). Then remove the number from the list. Now there remains m-1 numbers, so we call rand.nextInt(m-1). The number we get represents the position in the list. If it is less than the first number, then it is the second number, since the part of list prior to the first number wasn't changed by the removal of the first number. If the position is greater than or equal to the first number, the second number is position+1. Do some further derivation, you can get this algorithm.

Explanation

This algorithm has O(n^2) complexity. So it is good for generating small amount of unique numbers from a large set. While the shuffle based algorithm need at least O(m) to do the shuffle.

Also shuffle based algorithm need memory to store every possible outcome to do the shuffle, this algorithm doesn’t need.

Answer 2

I have come here from another question, which has been duplicate of this question (Generating unique random number in java)

1. Store 1 to 100 numbers in an Array.

2. Generate random number between 1 to 100 as position and return array[position-1] to get the value

3. Once you use a number in array, mark the value as -1 ( No need to maintain another array to check if this number is already used)

4. If value in array is -1, get the random number again to fetch new location in array.

Answer 3

This isn't significantly different from other answers, but I wanted the array of integers in the end:

    Integer[] indices = new Integer[n];
    Arrays.setAll(indices, i -> i);
    Collections.shuffle(Arrays.asList(indices));
    return Arrays.stream(indices).mapToInt(Integer::intValue).toArray();

Answer 4

• Add each number in the range sequentially in a list structure.
• Shuffle it.
• Take the first 'n'.

Here is a simple implementation. This will print 3 unique random numbers from the range 1-10.

import java.util.ArrayList;
import java.util.Collections;

public class UniqueRandomNumbers {

    public static void main(String[] args) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        for (int i=1; i<11; i++) {
            list.add(new Integer(i));
        }
        Collections.shuffle(list);
        for (int i=0; i<3; i++) {
            System.out.println(list.get(i));
        }
    }
}

---

The first part of the fix with the original approach, as Mark Byers pointed out in an answer now deleted, is to use only a single Random instance.

That is what is causing the numbers to be identical. A Random instance is seeded by the current time in milliseconds. For a particular seed value, the 'random' instance will return the exact same sequence of pseudo random numbers.

NOTE that the public Integer​(int value) constructor is deprecated since Java 9.

The first for loop can simply be changed to:

for (int i = 1; i < 11; i++) {
  list.add(i);
}

Answer 5

I feel like this method is worth mentioning.

   private static final Random RANDOM = new Random();    
   /**
     * Pick n numbers between 0 (inclusive) and k (inclusive)
     * While there are very deterministic ways to do this,
     * for large k and small n, this could be easier than creating
     * an large array and sorting, i.e. k = 10,000
     */
    public Set<Integer> pickRandom(int n, int k) {
        final Set<Integer> picked = new HashSet<>();
        while (picked.size() < n) {
            picked.add(RANDOM.nextInt(k + 1));
        }
        return picked;
    }

Answer 6

You can generate n unique random number between 0 to n-1 in java

public static void RandomGenerate(int n)
{
     Set<Integer> st=new HashSet<Integer>();
     Random r=new Random();
     while(st.size()<n)
     {
        st.add(r.nextInt(n));
     }

}