Creating random numbers with no duplicates

user_476438 0
Viewed 327721

In this case, the MAX is only 5, so I could check the duplicates one by one, but how could I do this in a simpler way? For example, what if the MAX has a value of 20? Thanks.

int MAX = 5;

for (i = 1 , i <= MAX; i++)
{
        drawNum[1] = (int)(Math.random()*MAX)+1;

        while (drawNum[2] == drawNum[1])
        {
             drawNum[2] = (int)(Math.random()*MAX)+1;
        }
        while ((drawNum[3] == drawNum[1]) || (drawNum[3] == drawNum[2]) )
        {
             drawNum[3] = (int)(Math.random()*MAX)+1;
        }
        while ((drawNum[4] == drawNum[1]) || (drawNum[4] == drawNum[2]) || (drawNum[4] == drawNum[3]) )
        {
             drawNum[4] = (int)(Math.random()*MAX)+1;
        }
        while ((drawNum[5] == drawNum[1]) ||
               (drawNum[5] == drawNum[2]) ||
               (drawNum[5] == drawNum[3]) ||
               (drawNum[5] == drawNum[4]) )
        {
             drawNum[5] = (int)(Math.random()*MAX)+1;
        }

}
random java
20 Answers
//random numbers are 0,1,2,3 
ArrayList<Integer> numbers = new ArrayList<Integer>();   
Random randomGenerator = new Random();
while (numbers.size() < 4) {

    int random = randomGenerator .nextInt(4);
    if (!numbers.contains(random)) {
        numbers.add(random);
    }
}
user_10455460

This would be a lot simpler in java-8:

Stream.generate(new Random()::ints)
            .flatMap(IntStream::boxed)
            .distinct()
            .limit(16) // whatever limit you might need
            .toArray(Integer[]::new);
user_10593720

Instead of doing all this create a LinkedHashSet object and random numbers to it by Math.random() function .... if any duplicated entry occurs the LinkedHashSet object won't add that number to its List ... Since in this Collection Class no duplicate values are allowed .. in the end u get a list of random numbers having no duplicated values .... :D

user_5758650

With Java 8 upwards you can use the ints method from the IntStream interface:

Returns an effectively unlimited stream of pseudorandom int values.

Random r = new Random();
int randomNumberOrigin = 0;
int randomNumberBound = 10;
int size = 5;
int[] unique = r.ints(randomNumberOrigin, randomNumberBound)
                .distinct()
                .limit(size)
                .toArray();
user_13668710

Following code create a sequence random number between [1,m] that was not generated before.

public class NewClass {

    public List<Integer> keys = new ArrayList<Integer>();

    public int rand(int m) {
        int n = (int) (Math.random() * m + 1);
        if (!keys.contains(n)) {
            keys.add(n);
            return n;
        } else {
            return rand(m);
        }
    }

    public static void main(String[] args) {
        int m = 4;
        NewClass ne = new NewClass();
        for (int i = 0; i < 4; i++) {
            System.out.println(ne.rand(m));
        }
        System.out.println("list: " + ne.keys);
    }
}
user_47212650

The most easy way is use nano DateTime as long format. System.nanoTime();

user_65502620

I created a snippet that generates no duplicate random integer. the advantage of this snippet is that you can assign the list of an array to it and generate the random item, too.

No duplication random generator class

user_80406970

With Java 8 using the below code, you can create 10 distinct random Integer Numbers within a range of 1000.

Random random = new Random();
Integer[] input9 = IntStream.range(1, 10).map(i -> random.nextInt(1000)).boxed().distinct()
                .toArray(Integer[]::new);
System.out.println(Arrays.toString(input9));

Modify the range to generate more numbers example : range(1,X). It will generate X distinct random numbers.

Modify the nextInt value to select the random number range : random.nextInt(Y)::random number will be generated within the range Y

user_28308420
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