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

user42155 Source

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

I have tried the following, but those do not work:

Attempt 1:

/ Bug: `randomNum` can be bigger than `maximum`.
randomNum = minimum + (int)(Math.random() * maximum);

Attempt 2:

// Bug: `randomNum` can be smaller than `minimum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;


answered 9 years ago krosenvold #1


minimum + rn.nextInt(maxValue - minvalue + 1)

answered 9 years ago Greg Case #2

In Java 1.7 or later, the standard way to do this 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 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. In those situations, the pre-Java 1.7 technique shown below 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.

answered 9 years ago Michael Myers #3

 rand.nextInt((max+1) - min) + min;

answered 9 years ago Chinnery #4

I wonder if any of the random number generating methods provided by an Apache Commons Math library would fit the bill.

For example: RandomDataGenerator.nextInt or RandomDataGenerator.nextLong

answered 9 years ago user2427 #5

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

Or take a look to RandomUtils from Apache Commons.

answered 9 years ago Bill the Lizard #6

You can edit your second code example to:

Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum =  rn.nextInt(range) + minimum;

answered 9 years ago TJ_Fischer #7

Note that this approach is more biased and less efficient than a nextInt approach,

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))

answered 9 years ago raupach #8

When you need a lot of random numbers, I do not recommend the Random class in the API. It has just a too small period. Try the Mersenne twister instead. There is a Java implementation.

answered 9 years ago Matt R #9

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

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 (

answered 9 years ago jackson #10


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.

answered 8 years ago sam #11

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);

answered 8 years ago ganesh #12

rand.nextInt((max+1) - min) + min;

This is working fine.

answered 7 years ago Joel Sjöstrand #13

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 naïve 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.

answered 7 years ago AZ_ #14

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);

answered 6 years ago gerardw #15

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);

answered 6 years ago Garrett Hall #16

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;

answered 6 years ago Hospes #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);


  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){

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

answered 6 years ago Luke Taylor #18

This methods might be convenient to use:

This method will return a random number between the provided min and max 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 min and max value (so the generated number could also be the min or max number):

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

    return randomNumber;

answered 5 years ago sachit #19

You can use this code snippet which will resolve your problem:

Random r = new Random();
int myRandomNumber = 0;
myRandomNumber = r.nextInt(maxValue-minValue+1)+minValue;

Use myRandomNumber (which will give you a number within a range).

answered 5 years ago jatin3893 #20

I just generate a random number using Math.random() and multiply it by a big number, let's say 10000. So, I get a number between 0 to 10,000 and call this number i. Now, if I need numbers between (x, y), then do the following:

i = x + (i % (y - x));

So, all i's are numbers between x and y.

To remove the bias as pointed out in the comments, rather than multiplying it by 10000 (or the big number), multiply it by (y-x).

answered 5 years ago Arun Abraham #21

One of my friends had asked me this same question in university today (his requirements was to generate a random number between 1 & -1). So I wrote this, and it works fine so far with my testing. There are ideally a lot of ways to generate random numbers given a range. Try this:


private static float getRandomNumberBetween(float numberOne, float numberTwo) throws Exception{

    if (numberOne == numberTwo){
        throw new Exception("Both the numbers can not be equal");

    float rand = (float) Math.random();
    float highRange = Math.max(numberOne, numberTwo);
    float lowRange = Math.min(numberOne, numberTwo);

    float lowRand = (float) Math.floor(rand-1);
    float highRand = (float) Math.ceil(rand+1);

    float genRand = (highRange-lowRange)*((rand-lowRand)/(highRand-lowRand))+lowRange;

    return genRand;

Execute like this:

System.out.println( getRandomNumberBetween(1,-1));

answered 5 years ago andrew #22

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)

answered 5 years ago agpt #23

Try using org.apache.commons.lang.RandomStringUtils class. Yes, it sometimes give a repeated number adjacently, but it will give the value between 5 and 15:

    while (true)
        int abc = Integer.valueOf(RandomStringUtils.randomNumeric(1));
        int cd = Integer.valueOf(RandomStringUtils.randomNumeric(2));
        if ((cd-abc) >= 5 && (cd-abc) <= 15)

answered 5 years ago Abel Melquiades Callejo #24

If you want to try the answer with the most votes above, you can simply use this code:

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

    public static void main(String[] args)

It is just clean and simple.

answered 4 years ago Akash Malhotra #25

I think this code will work for it. Please try this:

import java.util.Random;
public final class RandomNumber {

    public static final void main(String... aArgs) {
        log("Generating 10 random integers in range 1..10.");
        int START = 1;
        int END = 10;
        Random randomGenerator = new Random();
        for (int idx=1; idx<=10; ++idx) {

            // int randomInt=randomGenerator.nextInt(100);
            // log("Generated : " + randomInt);

    private static void log(String aMessage) {

    private static void showRandomInteger(int aStart, int aEnd, Random aRandom) {
        if (aStart > aEnd) {
            throw new IllegalArgumentException("Start cannot exceed End.");
        long range = (long)aEnd - (long)aStart + 1;
        long fraction = (long) (range * aRandom.nextDouble());
        int randomNumber = (int) (fraction + aStart);
        log("Generated" + randomNumber);

answered 4 years ago user591593 #26

I am thinking to linearly normalize the generated random numbers into desired range by using the following. Let x be a random number, let a and b be the minimum and maximum range of desired normalized number.

Then below is just a very simple code snipplet to test the range produced by the linear mapping.

public static void main(String[] args) {
    int a = 100;
    int b = 1000;
    int lowest = b;
    int highest = a;
    int count = 100000;
    Random random = new Random();
    for (int i = 0; i < count; i++) {
        int nextNumber = (int) ((Math.abs(random.nextDouble()) * (b - a))) + a;
        if (nextNumber < a || nextNumber > b) {
            System.err.println("number not in range :" + nextNumber);
        else {
        if (nextNumber < lowest) {
            lowest = nextNumber;
        if (nextNumber > highest) {
            highest = nextNumber;
    System.out.println("Produced " + count + " numbers from " + lowest
            + " to " + highest);

answered 4 years ago Prof Mo #27

Just use the Random class:

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

answered 4 years ago Jorge #28

I will simply state what is wrong with the solutions provided by the question and why the errors.

Solution 1:

randomNum = minimum + (int)(Math.random()*maximum); 

Problem: randomNum is assigned values numbers bigger than maximum.

Explanation: Suppose our minimum is 5, and your maximum is 10. Any value from Math.random() greater than 0.6 will make the expression evaluate to 6 or greater, and adding 5 makes it greater than 10 (your maximum). The problem is you are multiplying the random number by the maximum (which generates a number almost as big as the maximum) and then adding the minimum. Unless the minimum is 1, it's not correct. You have to switch to, as mentioned in other answers

randomNum = minimum + (int)(Math.random()*(maximum-minimum+1))

The +1 is because Math.random() will never return 1.0.

Solution 2:

Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;

Your problem here is that '%' may return a negative number if the first term is smaller than 0. Since rn.nextInt() returns negative values with ~50% chance, you will also not get the expected result.

This, was, however, almost perfect. You just had to look a bit further down the Javadoc, nextInt(int n). With that method available, doing

Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt(n);
randomNum =  minimum + i;

Would also return the desired result.

answered 4 years ago fedorp1 #29

You can either use the Random class to generate a random number and then use the .nextInt(maxNumber) to generate a random number. The maxNumber is the number that you want the maximum when generating a random number. Please remember though, that the Random class gives you the numbers 0 through maxNumber-1.

Random r = new Random();
int i = r.nextInt();

Another way to do this is to use the Math.Random() class which many classes in schools require you to use because it is more efficient and you don't have to declare a new Random object. To get a random number using Math.Random() you type in:

Math.random() * (max - min) + min;

answered 4 years ago ybn #30

import java.util.Random; 

public class RandomUtil {
    // Declare as class variable so that it is not re-seeded every call
    private static Random random = new Random();

     * Returns a psuedo-random number between min and max (both inclusive)
     * @param min Minimim value
     * @param max Maximim value. Must be greater than min.
     * @return Integer between min and max (both inclusive)
     * @see java.util.Random#nextInt(int)
    public static int nextInt(int min, int max) {
        // nextInt is normally exclusive of the top value,
        // so add 1 to make it inclusive
        return random.nextInt((max - min) + 1) + min;

answered 4 years ago Sunil Chawla #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);
    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.

answered 3 years ago Alexis C. #32

With 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.

Hope it helps! :)

answered 3 years ago Yakiv Mospan #33

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


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;

answered 3 years ago Muhammad Aamir Talib #34

private static Random random = new Random();    

public static int getRandomInt(int min, int max){
  return random.nextInt(max - min + 1) + min;


public static int getRandomInt(Random random, int min, int max)
  return random.nextInt(max - min + 1) + min;

answered 3 years ago hexabunny #35

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 class: Random

answered 3 years ago grep #36

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;

answered 3 years ago thangaraj #37

You can do something like this:

import java.awt.*;
import java.util.*;
import java.math.*;

public class Test {

    public static void main(String[] args) {
        int first, second;

        Scanner myScanner = new Scanner(;

        System.out.println("Enter first integer: ");
        int numOne;
        numOne = myScanner.nextInt();
        System.out.println("You have keyed in " + numOne);

        System.out.println("Enter second integer: ");
        int numTwo;
        numTwo = myScanner.nextInt();
        System.out.println("You have keyed in " + numTwo);

        Random generator = new Random();
        int num = (int)(Math.random()*numTwo);
        System.out.println("Random number: " + ((num>numOne)?num:numOne+num));

answered 3 years ago gifpif #38

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;

answered 2 years ago Andrey #39

Random number from the range [min..max] inclusive:

int randomFromMinToMaxInclusive = ThreadLocalRandom.current()
        .nextInt(min, max + 1);

answered 1 year ago Hiren Patel #40

This will generate Random numbers list with range (Min - Max) with no duplicate.

generateRandomListNoDuplicate(1000, 8000, 500);

Add this method.

private void generateRandomListNoDuplicate(int min, int max, int totalNoRequired) {
    Random rng = new Random();
    Set<Integer> generatedList = new LinkedHashSet<>();
    while (generatedList.size() < totalNoRequired) {
        Integer radnomInt = rng.nextInt(max - min + 1) + min;

Hope this will help you.

answered 1 year ago user_3380739 #41

Most of above suggestion don't consider 'overflow', For example: min = Integer.MIN_VALUE, max = 100. One of correct approaches I have so far is:

final long mod = max- min + 1L;
final int next = (int) (Math.abs(rand.nextLong() % mod) + min);

answered 1 year ago false9striker #42

You could use

RandomStringUtils.randomNumeric(int count)

method also which is from apache commons.

answered 1 year ago firephil #43

A different approach using Java 8 IntStream and Collections.shuffle

import java.util.ArrayList;
import java.util.Collections;

public class Main {

    public static void main(String[] args) {

        IntStream range = IntStream.rangeClosed(5,10);
        ArrayList<Integer> ls =  new ArrayList<Integer>();

        //populate the ArrayList
        range.forEach(i -> ls.add(new Integer(i)) );

        //perform a random shuffle  using the Collections Fisher-Yates shuffle

The equivalent in Scala

import scala.util.Random

object RandomRange extends App{
  val x =  Random.shuffle(5 to 10)

answered 1 year ago Nepster #44

I have created a method to get a unique integer in a given range.

      * minNum is the minimum possible random number
      * maxNum is the maximum possible random number
      * numbersNeeded is the quantity of random number required
      * the give method provides you with unique random number between min & max range
public static Set<Integer> getUniqueRandomNumbers( int minNum , int maxNum ,int numbersNeeded ){

    if(minNum >= maxNum)
        throw new IllegalArgumentException("maxNum must be greater than minNum");

    if(! (numbersNeeded > (maxNum - minNum + 1) ))
        throw new IllegalArgumentException("numberNeeded must be greater then difference b/w (max- min +1)");

    Random rng = new Random(); // Ideally just create one instance globally

    // Note: use LinkedHashSet to maintain insertion order
    Set<Integer> generated = new LinkedHashSet<Integer>();
    while (generated.size() < numbersNeeded)
        Integer next = rng.nextInt((maxNum - minNum) + 1) + minNum;

        // As we're adding to a set, this will automatically do a containment check
    return generated;

answered 11 months ago Jake O #45

Using Java 8 Streams,

  • Pass the initialCapacity - How many numbers
  • Pass the randomBound - from x to randomBound
  • Pass true/false for sorted or not
  • Pass an new Random() object


public static List<Integer> generateNumbers(int initialCapacity, int randomBound, Boolean sorted, Random random) {

    List<Integer> numbers = random.ints(initialCapacity, 1, randomBound).boxed().collect(Collectors.toList());

    if (sorted)

    return numbers;

It generates numbers from 1-Randombound in this example.

answered 10 months ago Divyesh Kanzariya #46

If you already use Commons Lang API 2.x or latest version then there is one class for random number generation RandomUtils.

public static int nextInt(int n)

Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive), from the Math.random() sequence.

Parameters: n - the specified exclusive max-value

int random = RandomUtils.nextInt(1000000);

Note: In RandomUtils have many methods for random number generation

answered 10 months ago Simon W. #47

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. Hope it helps :)

answered 9 months ago Mulalo Madida #48

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);

answered 7 months ago Gihan Chathuranga #49

This is the easy way to do this.

import java.util.Random;
class Example{
    public static void main(String args[]){
        /*-To test-
        for(int i = 1 ;i<20 ; i++){

        int randomnumber = randomnumber();


    public static int randomnumber(){
        Random rand = new Random();
        int randomNum = rand.nextInt(6) + 5;

        return randomNum;

In there 5 is the starting point of random numbers. 6 is the range including number 5.

answered 5 months ago Sam2016 #50

A simple way to generate n random numbers between a and b e.g a =90, b=100, n =20

Random r = new Random();
for(int i =0; i<20; i++){
    System.out.println(r.ints(90, 100).iterator().nextInt());

r.ints() returns an IntStream and has several useful methods, have look at its API.

answered 5 months ago namezhouyu #51

Random random = new Random();
  int max = 10;
  int min = 3;
int randomNum = random.nextInt(max) % (max - min + 1) + min;

answered 5 months ago Lawakush Kurmi #52

To generate random number "in between two numbers" use 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.

answered 4 months ago Icy TV #53

I didn't read through all of it, but did anyone say:

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

answered 4 months ago ledlogic #54

There is a library at for handling selection of ranges.

StochIntegerSelector randomIntegerSelector = new StochIntegerSelector();
Integer selectInteger = randomIntegerSelector.selectInteger();

It has edge inclusion/preclusion.

answered 2 months ago BMAM #55

The below code generates a random number between 100,000 and 900,000. This code will generate six digit values. I'm using this code to generate a six-digit OTP.

Use import java.util.Random to use this random method.

import java.util.Random;

// Six digits random number generation for OTP
Random rnd = new Random();
long longregisterOTP = 100000 + rnd.nextInt(900000);

answered 3 weeks ago Siddhartha Thota #56

It's pretty straightforward answer.

-> Say you want range between 0-9, 0 is min and 9 is max. The below function will print anything between 0 and 9. It's same for all ranges.

public static void main(String[] args) {
    int b = randomNumberRange(0, 9);
    int d = randomNumberRange (100, 200);
    System.out.println("value of b is " + b);
    System.out.println("value of d is " + d);

public static int randomNumberRange(int min, int max) {
int n = (max + 1 - min) + min;
return (int) (Math.random() * n);

comments powered by Disqus