How to generate 6 different random numbers in java

Question

I want to generate 6 different random numbers by using Math.random and store them into an array. How can I make sure that they are different? I know I need to use for-loop t

Answer 1

Create a variable last; initialize it to 0.

Next, in a loop x from 0 to 5, create a random number between last+1 and 49-6+x. Store this number in a list, and set last to the number generated this way.

You will end up with an ordered list of 6 random numbers in the range of 1..49 with no repeats.

Answer 2

In Java 8:

final int[] ints = new Random().ints(1, 50).distinct().limit(6).toArray();

In Java 7:

public static void main(final String[] args) throws Exception {
    final Random random = new Random();
    final Set<Integer> intSet = new HashSet<>();
    while (intSet.size() < 6) {
        intSet.add(random.nextInt(49) + 1);
    }
    final int[] ints = new int[intSet.size()];
    final Iterator<Integer> iter = intSet.iterator();
    for (int i = 0; iter.hasNext(); ++i) {
        ints[i] = iter.next();
    }
    System.out.println(Arrays.toString(ints));
}

Just a little messier. Not helped by the fact that it's pretty tedious to unbox the Set<Integer> into an int[].

It should be noted that this solution should be fine of the number of required values is significantly smaller than the range. As 1..49 is quite a lot larger than 6 you're fine. Otherwise performance rapidly degrades.

Answer 3

Instead of checking that the array has no duplicates, you can use a bit more smartness while generating the numbers, such that uniqueness is enforced at the outset.

1. Create a boolean[] as long as your range (49 entries);
2. generate a random number from the full range;
3. put that number into your output array;
4. "cross out" the corresponding index in the boolean[];
5. now generate another random number, but curtail the range by one (now 48);
6. instead of directly using that number as output, scan your boolean[], counting all the non-crossed entries. Stop when you reach the count equal to the random number generated in step 5. The number corresponding to that entry is your output number;
7. go to step 4.

Answer 4

Generate any 6 numbers (not necessarily different). Order them.

a1 <= a2 <= a3 <= a4 <= a5 <= a6

Now take these 6 numbers

a1 < a2 + 1 < a3 + 2 < a4 + 3 < a5 + 4 < a6 + 5

These 6 are different and random.

The idea of this construct comes from some combinatorial proofs.

Its advantage is that it's simple, fast, and deterministic.
I think the time complexity is O(count*log(count)).
I wonder if it can be improved.

import java.util.TreeMap;

public class Test005 {

    public static void main(String[] args) {
        int count = 6;
        int min = 1;
        int max = 49;

        // random number mapped to the count of its occurrences
        TreeMap<Integer, Integer> mp = new TreeMap<Integer, Integer>();
        for (int i=0; i<count; i++){
             int d = ( min + (int) (Math.random() * (max-count+1)) );
             if (!mp.containsKey(d)){
                 mp.put(d, 0);
             }
             mp.put(d, mp.get(d) + 1);
        }

        // now ensure the output numbers are different
        int j = 0;
        for (int num : mp.keySet()){
            int cnt = mp.get(num);
            for (int i=0; i<cnt; i++){
                System.out.println(num + j);
                j++;
            }
        }
    }

}