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

I've just came up with a small idea for Java 8-.

Set<Integer> set = new LinkedHashSet<>();
while(set.size() != 6)
    set.add(rnd.nextInt(49) + 1);

Answer 2

You can use a Set.

Set<Integer> s = new HashSet<>();
while(s.size() != 6){
   s.add(1 + (int) (Math.random() * 49));
}

Integer[] arr = s.toArray(new Integer[s.size()]);

This is enough to do this in your case because the number of distinct random numbers is relatively small compared to the size of the range you generate them.

Otherwise I would go with @JBNizet approach.

Answer 3

That code generate numbers from 6 to 0 and save in ArrayList.

If generated number was duplicated the program generate numbers again.

If generated number is different that number is added.

Code:

private ArrayList<Integer> arraylist = new ArrayList<Integer>();

private Random rand = new Random();

public void insertNumber() {
    while (true) {
        int i = generateNumber();
        if(!isGenerateNumberExists(i)){
            addNumber(i);
            break;
        }
    }
}
//Generate numbers
private int generateNumber() {
    return rand.nextInt(6);
}
//Confirm if that number exists
private boolean isGenerateNumberExists(int y) {
    for (int num : arraylist) {
        if (num == y) {
            return true;
        }
    }
    return false;
}
//Add number to arrayList
private void addNumber(int x) {
    arraylist.add(x);
}

Answer 4

Just keep generating numbers and adding them to the array as long as they are unique; psuedocode:

num = genNextRand()

For (array length)
    If (num not in array)
        addToArray()

Repeat while length not equal 6

Answer 5

in your case n=6

     public static int[] chooseAny(int n){
        int[] lottery = new int[n];
        int[] chooseFrom = new int[49];
        for(int i=1 ; i <= 49 ; i++)
            chooseFrom[i-1] = i;
        Random rand = new Random();
        int N = 49;
        int index;
        for(int i=0 ; i < n ; i++){
            //pick random index
            index = rand.nextInt(N);
            lottery[i] = chooseFrom[index];
            chooseFrom[index] = chooseFrom[N-1];
            N--;
        }
        return lottery;
    }

Answer 6

Create a list containing the numbers 1 to 49.

Create a random number x between 0 and the size of the list, take the number being at index x in the list, and remove it from the list.

Repeat the previous step 5 times. And you're done. Note that java.util.Random has a nextInt(int max) method that you should use instead of Math.random().

Note regarding performance: this solution has an advantage compared to the "try until you get 6 different numbers" various solutions: it runs in a O(n) time. It doesn't matter much for 6 unique numbers out of 50, but if you want to get 48 or 49 unique random numbers out of 50, you'll start seeing a difference, because you might have to generate many random numbers before getting one that isn't already in the set.

EDIT:

to reduce the cost induced by the removal of the elements in the list, you could instead simply replace the element at index x with the last element of the list (and at the second iteration, with the element at size - 2, etc.)