Question

How do I generate random integers within a specific range in Java?

How do I generate a random int value in a specific range?

The following methods have bugs related to integer overflow:

randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
 4108  5249164  4108
1 Jan 1970

Solution

 4383

In Java 1.7 or later, the standard way to do this (generate a basic non-cryptographically secure random integer in the range [min, max]) is as follows:

import java.util.concurrent.ThreadLocalRandom;

// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);

See the relevant JavaDoc. This approach has the advantage of not needing to explicitly initialize a java.util.Random instance, which can be a source of confusion and error if used inappropriately.

However, conversely with ThreadLocalRandom there is no way to explicitly set the seed so it can be difficult to reproduce results in situations where that is useful such as testing or saving game states or similar.

As of Java 17, the psuedorandom number generating classes in the standard library implement the RandomGenerator interface. See the linked JavaDoc for more information. For example, if a cryptographically strong random number generator is desired, the SecureRandom class can be used.

Before Java 1.7, the standard way to do this is as follows:

import java.util.Random;

/**
 * Returns a pseudo-random number between min and max, inclusive.
 * The difference between min and max can be at most
 * <code>Integer.MAX_VALUE - 1</code>.
 *
 * @param min Minimum value
 * @param max Maximum value.  Must be greater than min.
 * @return Integer between min and max, inclusive.
 * @see java.util.Random#nextInt(int)
 */
public static int randInt(int min, int max) {

    // NOTE: This will (intentionally) not run as written so that folks
    // copy-pasting have to think about how to initialize their
    // Random instance.  Initialization of the Random instance is outside
    // the main scope of the question, but some decent options are to have
    // a field that is initialized once and then re-used as needed or to
    // use ThreadLocalRandom (if using at least Java 1.7).
    // 
    // In particular, do NOT do 'Random rand = new Random()' here or you
    // will get not very good / not very random results.
    Random rand;

    // nextInt is normally exclusive of the top value,
    // so add 1 to make it inclusive
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return randomNum;
}

See the relevant JavaDoc. In practice, the java.util.Random class is often preferable to java.lang.Math.random().

In particular, there is no need to reinvent the random integer generation wheel when there is a straightforward API within the standard library to accomplish the task.

2008-12-12
Greg Case

Solution

 1513

Note that this approach is more biased and less efficient than a nextInt approach, https://stackoverflow.com/a/738651/360211

One standard pattern for accomplishing this is:

Min + (int)(Math.random() * ((Max - Min) + 1))

The Java Math library function Math.random() generates a double value in the range [0,1). Notice this range does not include the 1.

In order to get a specific range of values first, you need to multiply by the magnitude of the range of values you want covered.

Math.random() * ( Max - Min )

This returns a value in the range [0,Max-Min), where 'Max-Min' is not included.

For example, if you want [5,10), you need to cover five integer values so you use

Math.random() * 5

This would return a value in the range [0,5), where 5 is not included.

Now you need to shift this range up to the range that you are targeting. You do this by adding the Min value.

Min + (Math.random() * (Max - Min))

You now will get a value in the range [Min,Max). Following our example, that means [5,10):

5 + (Math.random() * (10 - 5))

But, this still doesn't include Max and you are getting a double value. In order to get the Max value included, you need to add 1 to your range parameter (Max - Min) and then truncate the decimal part by casting to an int. This is accomplished via:

Min + (int)(Math.random() * ((Max - Min) + 1))

And there you have it. A random integer value in the range [Min,Max], or per the example [5,10]:

5 + (int)(Math.random() * ((10 - 5) + 1))
2008-12-12
TJ_Fischer

Solution

 463

Use:

Random ran = new Random();
int x = ran.nextInt(6) + 5;

The integer x is now the random number that has a possible outcome of 5-10.

2009-09-04
jackson

Solution

 205

Use:

minValue + rn.nextInt(maxValue - minValue + 1)
2008-12-12
krosenvold

Solution

 202

With Java 8 they introduced the method ints(int randomNumberOrigin, int randomNumberBound) in the Random class.

For example if you want to generate five random integers (or a single one) in the range [0, 10], just do:

Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();

The first parameter indicates just the size of the IntStream generated (which is the overloaded method of the one that produces an unlimited IntStream).

If you need to do multiple separate calls, you can create an infinite primitive iterator from the stream:

public final class IntRandomNumberGenerator {

    private PrimitiveIterator.OfInt randomIterator;

    /**
     * Initialize a new random number generator that generates
     * random numbers in the range [min, max]
     * @param min - the min value (inclusive)
     * @param max - the max value (inclusive)
     */
    public IntRandomNumberGenerator(int min, int max) {
        randomIterator = new Random().ints(min, max + 1).iterator();
    }

    /**
     * Returns a random number in the range (min, max)
     * @return a random number in the range (min, max)
     */
    public int nextInt() {
        return randomIterator.nextInt();
    }
}

You can also do it for double and long values.

2014-11-26
Alexis C.

Solution

 123

You can edit your second code example to:

Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum =  rn.nextInt(range) + minimum;
2008-12-12
Bill the Lizard

Solution

 110

Just a small modification of your first solution would suffice.

Random rand = new Random();
randomNum = minimum + rand.nextInt((maximum - minimum) + 1);

See more here for implementation of Random

2015-03-12
hexabunny

Solution

 102

ThreadLocalRandom equivalent of class java.util.Random for multithreaded environment. Generating a random number is carried out locally in each of the threads. So we have a better performance by reducing the conflicts.

int rand = ThreadLocalRandom.current().nextInt(x,y);

x,y - intervals e.g. (1,10)

2013-02-12
andrew

Solution

 79

The Math.Random class in Java is 0-based. So, if you write something like this:

Random rand = new Random();
int x = rand.nextInt(10);

x will be between 0-9 inclusive.

So, given the following array of 25 items, the code to generate a random number between 0 (the base of the array) and array.length would be:

String[] i = new String[25];
Random rand = new Random();
int index = 0;

index = rand.nextInt( i.length );

Since i.length will return 25, the nextInt( i.length ) will return a number between the range of 0-24. The other option is going with Math.Random which works in the same way.

index = (int) Math.floor(Math.random() * i.length);

For a better understanding, check out forum post Random Intervals (archive.org).

2009-01-08
Matt R

Solution

 64

It can be done by simply doing the statement:

Randomizer.generate(0, 10); // Minimum of zero and maximum of ten

Below is its source code.

File Randomizer.java

public class Randomizer {
    public static int generate(int min, int max) {
        return min + (int)(Math.random() * ((max - min) + 1));
    }
}

It is just clean and simple.

2013-09-01
Abel Callejo

Solution

 53

Forgive me for being fastidious, but the solution suggested by the majority, i.e., min + rng.nextInt(max - min + 1)), seems perilous due to the fact that:

  • rng.nextInt(n) cannot reach Integer.MAX_VALUE.
  • (max - min) may cause overflow when min is negative.

A foolproof solution would return correct results for any min <= max within [Integer.MIN_VALUE, Integer.MAX_VALUE]. Consider the following naive implementation:

int nextIntInRange(int min, int max, Random rng) {
   if (min > max) {
      throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
   }
   int diff = max - min;
   if (diff >= 0 && diff != Integer.MAX_VALUE) {
      return (min + rng.nextInt(diff + 1));
   }
   int i;
   do {
      i = rng.nextInt();
   } while (i < min || i > max);
   return i;
}

Although inefficient, note that the probability of success in the while loop will always be 50% or higher.

2011-01-10
Joel Sj&#246;strand

Solution

 34

I use this:

 /**
   * @param min - The minimum.
   * @param max - The maximum.
   * @return A random double between these numbers (inclusive the minimum and maximum).
   */
 public static double getRandom(double min, double max) {
   return (Math.random() * (max + 1 - min)) + min;
 }

You can cast it to an Integer if you want.

2017-05-28
Simon

Solution

 34

As of Java 7, you should no longer use Random. For most uses, the random number generator of choice is now ThreadLocalRandom.

For fork join pools and parallel streams, use SplittableRandom.

Joshua Bloch. Effective Java. Third Edition.

Starting from Java 8

For fork join pools and parallel streams, use SplittableRandom (it implements SplittableGenerator interface, see Java 17 notes below) that is usually faster, has a better statistical independence and uniformity properties in comparison with Random.

To generate a random int in the range [0, 1_000]:

int n = new SplittableRandom().nextInt(0, 1_001);

To generate a random int[100] array of values in the range [0, 1_000]:

int[] a = new SplittableRandom().ints(100, 0, 1_001).parallel().toArray();

To return a Stream of random values:

IntStream stream = new SplittableRandom().ints(100, 0, 1_001);

Java 17

It is recommended that multithreaded applications use either ThreadLocalRandom or (preferably) pseudorandom number generators that implement the RandomGenerator.SplittableGenerator or RandomGenerator.JumpableGenerator interface.

RandomGenerator1 documentation.


1 - RandomGenerator is a modern interface that was introduced in Java 17. It is a common protocol for objects that generate random or pseudorandom sequences of numbers and has more features and methods than Random.

2018-04-11
Oleksandr Pyrohov

Solution

 31
 rand.nextInt((max+1) - min) + min;
2008-12-12
Michael Myers

Solution

 31

Let us take an example.

Suppose I wish to generate a number between 5-10:

int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);

Let us understand this...

Initialize max with highest value and min with the lowest value.

Now, we need to determine how many possible values can be obtained. For this example, it would be:

5, 6, 7, 8, 9, 10

So, count of this would be max - min + 1.

i.e. 10 - 5 + 1 = 6

The random number will generate a number between 0-5.

i.e. 0, 1, 2, 3, 4, 5

Adding the min value to the random number would produce:

5, 6, 7, 8, 9, 10

Hence we obtain the desired range.

2014-08-03
Sunil Chawla

Solution

 29

Generate a random number for the difference of min and max by using the nextint(n) method and then add min number to the result:

Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
2015-05-27
gifpif

Solution

 21

These methods might be convenient to use:

This method will return a random number between the provided minimum and maximum value:

public static int getRandomNumberBetween(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt(max - min) + min;
    if (randomNumber == min) {
        // Since the random number is between the min and max values, simply add 1
        return min + 1;
    } else {
        return randomNumber;
    }
}

and this method will return a random number from the provided minimum and maximum value (so the generated number could also be the minimum or maximum number):

public static int getRandomNumberFrom(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt((max + 1) - min) + min;

    return randomNumber;
}
2012-08-12
Luke Taylor

Solution

 21

Just use the Random class:

Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
2013-12-24
Prof Mo

Solution

 21

To generate a random number "in between two numbers", use the following code:

Random r = new Random();
int lowerBound = 1;
int upperBound = 11;
int result = r.nextInt(upperBound-lowerBound) + lowerBound;

This gives you a random number in between 1 (inclusive) and 11 (exclusive), so initialize the upperBound value by adding 1. For example, if you want to generate random number between 1 to 10 then initialize the upperBound number with 11 instead of 10.

2017-11-02
Lawakush Kurmi

Solution

 20
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();

Or take a look to RandomUtils from Apache Commons.

2008-12-12
user2427

Solution

 20

In case of rolling a dice it would be random number between 1 to 6 (not 0 to 6), so:

face = 1 + randomNumbers.nextInt(6);
2010-02-16
sam

Solution

 19

You can achieve that concisely in Java 8:

Random random = new Random();

int max = 10;
int min = 5;
int totalNumber = 10;

IntStream stream = random.ints(totalNumber, min, max);
stream.forEach(System.out::println);
2017-06-20
user5150135

Solution

 18

Another option is just using Apache Commons:

import org.apache.commons.math.random.RandomData;
import org.apache.commons.math.random.RandomDataImpl;

public void method() {
    RandomData randomData = new RandomDataImpl();
    int number = randomData.nextInt(5, 10);
    // ...
 }
2012-01-18
gerardw

Solution

 18

Here's a helpful class to generate random ints in a range with any combination of inclusive/exclusive bounds:

import java.util.Random;

public class RandomRange extends Random {
    public int nextIncInc(int min, int max) {
        return nextInt(max - min + 1) + min;
    }

    public int nextExcInc(int min, int max) {
        return nextInt(max - min) + 1 + min;
    }

    public int nextExcExc(int min, int max) {
        return nextInt(max - min - 1) + 1 + min;
    }

    public int nextIncExc(int min, int max) {
        return nextInt(max - min) + min;
    }
}
2012-02-15
Garrett Hall

Solution

 17
public static Random RANDOM = new Random(System.nanoTime());

public static final float random(final float pMin, final float pMax) {
    return pMin + RANDOM.nextFloat() * (pMax - pMin);
}
2011-07-13
AZ_

Solution

 17

I found this example Generate random numbers :


This example generates random integers in a specific range.

import java.util.Random;

/** Generate random integers in a certain range. */
public final class RandomRange {

  public static final void main(String... aArgs){
    log("Generating random integers in the range 1..10.");

    int START = 1;
    int END = 10;
    Random random = new Random();
    for (int idx = 1; idx <= 10; ++idx){
      showRandomInteger(START, END, random);
    }

    log("Done.");
  }

  private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
    if ( aStart > aEnd ) {
      throw new IllegalArgumentException("Start cannot exceed End.");
    }
    //get the range, casting to long to avoid overflow problems
    long range = (long)aEnd - (long)aStart + 1;
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    int randomNumber =  (int)(fraction + aStart);    
    log("Generated : " + randomNumber);
  }

  private static void log(String aMessage){
    System.out.println(aMessage);
  }
} 

An example run of this class :

Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.
2012-06-07
Hospes

Solution

 15

Here is a simple sample that shows how to generate random number from closed [min, max] range, while min <= max is true

You can reuse it as field in hole class, also having all Random.class methods in one place

Results example:

RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert

Sources:

import junit.framework.Assert;
import java.util.Random;

public class RandomUtils extends Random {

    /**
     * @param min generated value. Can't be > then max
     * @param max generated value
     * @return values in closed range [min, max].
     */
    public int nextInt(int min, int max) {
        Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
        if (min == max) {
            return max;
        }

        return nextInt(max - min + 1) + min;
    }
}
2014-11-28
Yakiv Mospan

Solution

 15

It's better to use SecureRandom rather than just Random.

public static int generateRandomInteger(int min, int max) {
    SecureRandom rand = new SecureRandom();
    rand.setSeed(new Date().getTime());
    int randomNum = rand.nextInt((max - min) + 1) + min;
    return randomNum;
}
2015-03-26
grep

Solution

 15

The int nextInt(int origin, int bound) method was added in Java 17 as part of the RandomGenerator interface. This will generate a random integer in a given range:

// Returns a random int between minimum (inclusive) & maximum (exclusive)
int randomNum = RandomGenerator.getDefault().nextInt(minimum, maximum);

This interface is used for new random generation algorithms added in Java 17:

RandomGenerator.getDefault().nextInt(minimum, maximum);
RandomGenerator.of("L128X1024MixRandom").nextInt(minimum, maximum);
RandomGenerator.of("Xoroshiro128PlusPlus").nextInt(minimum, maximum);
// ...

The RandomGenerator interface was also added to the existing random generation classes (Random, SecureRandom, SplittableRandom, and ThreadLocalRandom). Therefore, as of Java 17, those four classes also have this bounded nextInt method:

new Random().nextInt(minimum, maximum);
new SecureRandom().nextInt(minimum, maximum);
new SplittableRandom().nextInt(minimum, maximum);
new ThreadLocalRandom().nextInt(minimum, maximum);

This method is new to Random and SecureRandom as of Java 17. Prior to Java 17, ThreadLocalRandom and SplittableRandom already had this method, though it was not specified by a shared interface.

2021-12-30
M. Justin