Generate more than one random numbers using native or built-in function

Question

I know for generating random number there is random function in java

for example 

       int randomNumber = ( int )( Math.random() * 9999 );

which will generate randomNumber from [0,9999] but it only returns one number by given range.

But I just want to know if there is any built in function or you can say native function which will generate more than one random numbers and it should also not match with each other

suppose from above example if i want to generate 4 number

it will return like 1,10,50,5544

Here you can see that there is a four random number and it is not matching with each other .

See [this answer](http://stackoverflow.com/questions/9423523/generate-unique-random-numbers-in-java?rq=1) for a possible solution — Michael Lang, Jul 11 '13 at 06:06
please take care of your tags, this here is a basic language question, completely unrelated to swing, java-me, java-ee or android (which I removed) — kleopatra, Jul 11 '13 at 06:12
As I get Java .. they try to provide only basic functions in the core of the language. Getting N unique results out of the Core Random function can be done in 2-3 rows of simple code .. so no point to implement this as a separate function. — d.raev, Jul 11 '13 at 06:15

score 4 · Answer 1 · answered Jul 11 '13 at 06:07

4

try something like this

Create an ArrayList and add your random number and while adding check if ArrayList already contains the number or not.

Eg:

ArrayList<Integer> numbers = new ArrayList<Integer>();
while (numbers.size()<=YOUR_MAX_SIZE)
{
    int randomInteger = ( int )( Math.random() * 9999 );
    if (!numbers.contains(randomInteger)) {
    {
       numbers.add(randomInteger);
    }
}

answered Jul 11 '13 at 06:07

Michael Shrestha

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Ya michal that's i know by that way i can do by different different way but i just want to know if there is built in native function or not! – Siddhpura Amit Jul 11 '13 at 06:10
I doubt there is any.. I haven't found any such methods.. but if there exists such method that will be very great... – Michael Shrestha Jul 11 '13 at 06:12
@SiddhpuraAmit - No, there's not. Check the documentation. – OptimusCrime Jul 11 '13 at 06:12
+1 for being the only other answer so far that accounts for 'no duplicates'. – Andrew Thompson Jul 11 '13 at 06:16
1

If you use this solution with a `Set` you can get rid of the `contains` check. – Michael Lang Jul 11 '13 at 06:16
Thanks Michael for suggestion me i give +1 :) – Siddhpura Amit Jul 11 '13 at 06:24

Andrew Thompson · Answer 2 · 2013-07-11T06:14:26.713

4

Another technique (besides the one shown by @MichaelShrestha) is to put the entire range of numbers in a collection, then shuffle it and take the numbers in the shuffled order.

It has the advantage that whereas the other method might spin though many (many) numbers to find a non-duplicate random value, this will only have to be run once. OTOH, for small collections of numbers, @Michael's technique might be faster.

edited Jul 11 '13 at 06:14

answered Jul 11 '13 at 06:08

Andrew Thompson

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Thank you so so so much Andrew :) Will give u +1 points for nice suggestion :) – Siddhpura Amit Jul 11 '13 at 06:25

score 3 · Answer 3 · edited May 23 '17 at 10:28

3

Try this, as proposed here:

int min = 1;
int max = 10000;

Random r = new Random();
int i1 = r.nextInt(max - min + 1) + min;

EDIT:

With no duplicates:

Random rng = new Random(); // Ideally just create one instance globally
int min = 1;
int max = 10000;
Set<Integer> generated = new LinkedHashSet<Integer>();
while (generated.size() < numbersNeeded)
{
    Integer next = r.nextInt(max - min + 1) + min;
    generated.add(next);
}

LINK

edited May 23 '17 at 10:28

Community

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1

answered Jul 11 '13 at 06:09

g00dy

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No, I was thanking you for showing me that I missed that part :-) The down-vote made me saaad :-D – g00dy Jul 11 '13 at 06:18
Thanks G00dy for suggestion me i give +1 :) – Siddhpura Amit Jul 11 '13 at 06:24

score 0 · Answer 4 · answered Jul 11 '13 at 06:12

0

There is not built in api. but you can use something like this

public static int[] getRandoms(int amount, int range){
    int[] result = new int[amount];
    ArrayList tempList = new ArrayList();
    for(int i = 0; i< range; i++)
       tempList.add(i);
    Collections.shuffle(tempList);
    for(int i = 0; i< amount; i++)
       result[i] = (int)tempList.get(i);
    return result;
}

Here amount is the number of randome numbers you want and the range is the maximum range of random number

answered Jul 11 '13 at 06:12

stinepike

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Ya thanks :) by this way is also useful but there is no any built in native function??? – Siddhpura Amit Jul 11 '13 at 06:14
I haven't found any so far. But if you find any please share with us :) – stinepike Jul 11 '13 at 06:15
2

*"no any built in native function???"* 'Built in' reads like 'lazy ass' here. Put it in a method and there is your 'built in' function. Also note that while one '?' indicates a question, 2 or more typically indicates a bozo. – Andrew Thompson Jul 11 '13 at 06:21
Thanks Stine for suggestion me i give +1 :) – Siddhpura Amit Jul 11 '13 at 06:24
Ok ok Sorry Dear Bro Andrew thanks for your nice suggestion i have to keep in mind this technical terms :) – Siddhpura Amit Jul 11 '13 at 06:35

score 0 · Answer 5 · answered Jul 11 '13 at 06:14

    ArrayList<Integer> rnumbers = new ArrayList<Integer>(); 
    for (int i = 0; i < 4; i++) {
        int randomNumber = (int) (Math.random() * 9999);
        while (rnumbers.contains(randomNumber)) {
            randomNumber = (int) (Math.random() * 9999);
        }
        rnumbers.add(randomNumber);
    }

score 0 · Answer 6 · answered Jul 11 '13 at 06:25

As HashSet contains unique object... You can have the unique random numbers.

Set uniqueRandomNumber = new HashSet();
 int i = 5 //number Of Random Number Required;
 for(int = 0; i<5; 1++){
    int randomNumber = ( int )( Math.random() * 9999 );
    uniqueRandomNumber.add(randomNumber);
 }

Generate more than one random numbers using native or built-in function

6 Answers6