generate any random number of any Length in Java

Question

How to generate any random number of any Length in Java? Like, to generate a number of width 3, it should be between 100 to 999. So how to code this?
In image below:

I want to solve second problem.
This is what I written for first problem:

public int getRandomNumberReturntypediff(int min,int max) {
    Random rand = new Random();

    int  n = rand.nextInt(1000) + 1;
    Integer.toString(n);

    System.out.println("get Random Number with return type STRING" + n);
    return n;
}

Suppose I take parameter length and return String. How can I do this?

Answer 1

We use String for create number and the convert it to a BigInteger. I think the the best way for handle large number is to use BigInteger but if you need String just return that.

static List<Character> NUMBERS_WITHOUT_ZERO = List.of('1', '2', '3', '4', '5', '6', '7', '8', '9');
static List<Character> NUMBERS_WITH_ZERO = List.of('0', '1', '2', '3', '4', '5', '6', '7', '8', '9');
private static final SecureRandom random = new SecureRandom();

public static BigInteger randomNumber(int length) {
    StringBuilder res = new StringBuilder(length);
    res.append(NUMBERS_WITHOUT_ZERO.get(random.nextInt(NUMBERS_WITHOUT_ZERO.size())));
    for (int i = 2; i <= length; i++)
        res.append(NUMBERS_WITH_ZERO.get(random.nextInt(NUMBERS_WITH_ZERO.size())));
    return new BigInteger(res.toString());
}

I hope this is what you want.

Answer 2

Since you already have the actual bounded random number generation logic in your question, I assume your struggling with the calculation of the bounds.

You can calculate the minimum and maximum values by using powers of 10:

• Minimum = 10 ^ (length - 1)
• Maximum = (10 ^ length) - 1

So, for example, for a number length of 3, this will give you a minimum of 100, and a maximum of 999.

Here's a complete example:

import java.util.Random;

public class Randoms {
    public static void main(String args[]) {
        System.out.println(generateRandom(3));
    }

    private static String generateRandom(int length) {
        int min = (int) Math.pow(10, length - 1);
        int max = (int) Math.pow(10, length); // bound is exclusive

        Random random = new Random();

        return Integer.toString(random.nextInt(max - min) + min);
    }
}

Answer 3

I think this should work:

int n = min + (int) (Math.random() * (max - min + 1));

Answer 4

public void generateRandomNumber(int length){
     int max= Math.pow(10,length)-1;
     int min=Math.pow(10,length-1);
     int range = max - min + 1;
     int randomNumber = (int) (Math.random() * range) + min;
}     

Math.random() returns a value between greater than equal to 0.0 but less than 1.0

Answer 5

You need this method,

public static int getRandNum(int min, int max) {
    return (int)(Math.random()*(max-min)) + min;
}

Like you asked in your post in image, this method returns a random number between min (inclusive) and max (exclusive).

Answer 6

My variant

 public class RandomGen {

    public int getRandom(int min, int max) {
    Random random = new Random();
    return random.nextInt((max - min) + 1) + min;
     }
   }

    public class App {
      public static void main(String[] args) {
      RandomGen value = new RandomGen();
      System.out.println(value.getRandom(300, 400));
     }
    }

Answer 7

Not sure if this is still relevant if I understood, the solution comes out of the box in java.

Random sr = new SecureRandom();
IntStream ints = sr.ints(3, 100, 999);
Iterator<Integer> it = ints.boxed().collect(Collectors.toSet()).iterator();

The iterator is just like any other so you can play with it at will. Let me know if this is not what you want.