如何在 Java 中生成特定范围内的随机整数？

Question

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`.

Answer 1

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.

Answer 2

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))

Answer 3

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 .

Answer 4

利用：

minValue + rn.nextInt(maxValue - minValue + 1)

Answer 5

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. I hope it helps! :)

Answer 6

You can edit your second code example to:

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

Answer 7

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

Answer 8

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 eg (1,10)

Answer 9

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) .

Answer 10

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.

Answer 11

Forgive me for being fastidious, but the solution suggested by the majority, ie, 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.

Answer 12

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

Answer 13

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.

Answer 14

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.

ie 10 - 5 + 1 = 6

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

ie 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.

Answer 15

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

Answer 16

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 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);

Answer 17

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);

Answer 18

Just use the Random class:

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

Answer 19

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;
}

Answer 20

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.

Answer 21

如果掷骰子，它将是 1 到 6（不是 0 到 6）之间的随机数，所以：

face = 1 + randomNumbers.nextInt(6);

Answer 22

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

或者看看Apache Commons的 RandomUtils 。

Answer 23

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);

Answer 24

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;
    }
}

Answer 25

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);
}

Answer 26

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);
    // ...
 }

Answer 27

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.

Answer 28

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 .

Answer 29

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;
    }
}

Answer 30

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;
}

Answer 31

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

这工作正常。

Answer 32

private static Random random = new Random();    

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

OR

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

Answer 33

Java 17 introduced the RandomGenerator interface, which provides a int nextInt(int origin, int bound) method to get a random integer in a range:

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

In addition to being used for the new random generation algorithms added in Java 17, this interface was added to the existing random generation classes ( Random , SecureRandom , SplittableRandom , and ThreadLocalRandom ). Therefore, as of Java 17, these classes 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.

Answer 34

Random rng = new Random();
int min = 3;
int max = 11;
int upperBound = max - min + 1; // upper bound is exclusive, so +1
int num = min + rng.nextInt(upperBound);
System.out.println(num);

Answer 35

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).

Answer 36

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.

Answer 37

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;
    }
}

Answer 38

A different approach using Java 8 IntStream and Collections.shuffle

import java.util.stream.IntStream;
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
        Collections.shuffle(ls);
        System.out.println(ls);
    }
}

The equivalent in Scala

import scala.util.Random

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

Answer 39

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 {
            System.out.println(nextNumber);
        }
        if (nextNumber < lowest) {
            lowest = nextNumber;
        }
        if (nextNumber > highest) {
            highest = nextNumber;
        }
    }
    System.out.println("Produced " + count + " numbers from " + lowest
            + " to " + highest);
}

Answer 40

You can do something like this:

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

public class Test {

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

        Scanner myScanner = new Scanner(System.in);

        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));
    }
}

Answer 41

You could use the

RandomStringUtils.randomNumeric(int count)

method which is also from Apache Commons .

Answer 42

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

Answer 43

可以使用以下代码：

ThreadLocalRandom.current().nextInt(rangeStart, rangeEndExclusive)

Answer 44

You can use the following way to do that:

int range = 10;
int min = 5
Random r = new Random();
int = r.nextInt(range) + min;

Answer 45

int randomNum = 5 + (int)(Math.random()*5);

范围 5-10

Answer 46

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 (yx).

Answer 47

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++){
            System.out.print(randomnumber()+",");
        }
        */

        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.

Answer 48

Say you want range between 0-9, 0 is minimum and 9 is maximum. The below function will print anything between 0 and 9. It's the 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);
}

Answer 49

Making the following change in Attempt 1 should do the work -

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

Check this for working code.

Answer 50

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:

Function:

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));

Answer 51

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);
            showRandomInteger(START,END,randomGenerator);
        }
        log("Done");
    }

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

Answer 52

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;
        generatedList.add(radnomInt);
    }
}

Hope this will help you.

Answer 53

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
        generated.add(next);
    }
    return generated;
}

Answer 54

If you already use Commons Lang API 3.x or latest version then there is one class for random number generation RandomUtils .

public static int nextInt(int startInclusive, int endExclusive)

Returns a random integer within the specified range.

Parameters:

startInclusive - the specified starting value

endExclusive - the specified end value

int random = RandomUtils.nextInt(999,1000000);

Note: In RandomUtils have many methods for random number generation

Answer 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);
System.out.println(longregisterOTP);

Answer 56

The following is another example using Random and forEach

int firstNum = 20;//Inclusive
int lastNum = 50;//Exclusive
int streamSize = 10;
Random num = new Random().ints(10, 20, 50).forEach(System.out::println);

Answer 57

Use java.util for Random for general use.

You can define your minimum and maximum range to get those results.

Random rand=new Random();
rand.nextInt((max+1) - min) + min;

Answer 58

int func(int max, int min){

      int range = max - min + 1;
      
      // Math.random() function will return a random no between [0.0,1.0).
      int res = (int) ( Math.random()*range)+min;

      return res;
}

Answer 59

Use Apache Lang3 Commons

Integer.parseInt(RandomStringUtils.randomNumeric(6, 6));

Min Value 100000 to Max Value 999999

Answer 60

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;

Answer 61

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

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

Answer 62

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

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

Answer 63

There is a library at https://sourceforge.net/projects/stochunit/ for handling selection of ranges.

StochIntegerSelector randomIntegerSelector = new StochIntegerSelector();
randomIntegerSelector.setMin(-1);
randomIntegerSelector.setMax(1);
Integer selectInteger = randomIntegerSelector.selectInteger();

It has edge inclusion/preclusion.

Answer 64

import java.util.Random;

public class RandomSSNTest {

    public static void main(String args[]) {
        generateDummySSNNumber();
    }


    //831-33-6049
    public static void generateDummySSNNumber() {
        Random random = new Random();

        int id1 = random.nextInt(1000);//3
        int id2 = random.nextInt(100);//2
        int id3 = random.nextInt(10000);//4

        System.out.print((id1+"-"+id2+"-"+id3));
    }

}

You can also use

import java.util.concurrent.ThreadLocalRandom;
Random random = ThreadLocalRandom.current();

However, this class doesn't perform well in a multi-threaded environment.

Answer 65

To avoid repeating what have been said multiple times, I am showing an alternative for those that need a cryptographically stronger pseudo-random number generator by using the SecureRandom class, which extends the class Random . From source one can read:

This class provides a cryptographically strong random number generator (RNG). A cryptographically strong random number minimally complies with the statistical random number generator tests specified in FIPS 140-2, Security Requirements for Cryptographic Modules, section 4.9.1. Additionally, SecureRandom must produce non-deterministic output. Therefore any seed material passed to a SecureRandom object must be unpredictable, and all SecureRandom output sequences must be cryptographically strong, as described in RFC 1750: Randomness Recommendations for Security.

A caller obtains a SecureRandom instance via the no-argument constructor or one of the getInstance methods:

 SecureRandom random = new SecureRandom();

Many SecureRandom implementations are in the form of a pseudo-random number generator (PRNG), which means they use a deterministic algorithm to produce a pseudo-random sequence from a true random seed. Other implementations may produce true random numbers, and yet others may use a combination of both techniques.

To generate a random number between a min and max values inclusive:

public static int generate(SecureRandom secureRandom, int min, int max) {
        return min + secureRandom.nextInt((max - min) + 1);
}

for a given a min (inclusive) and max (exclusive) values:

return min + secureRandom.nextInt((max - min));

A running code example:

public class Main {

    public static int generate(SecureRandom secureRandom, int min, int max) {
        return min + secureRandom.nextInt((max - min) + 1);
    }

    public static void main(String[] arg) {
        SecureRandom random = new SecureRandom();
        System.out.println(generate(random, 0, 2 ));
    }
}

Source such as stackoverflow , baeldung , geeksforgeeks provide comparisons between Random and SecureRandom classes.

From baeldung one can read:

The most common way of using SecureRandom is to generate int, long, float, double or boolean values:

int randomInt = secureRandom.nextInt();
long randomLong = secureRandom.nextLong();
float randomFloat = secureRandom.nextFloat();
double randomDouble = secureRandom.nextDouble();
boolean randomBoolean = secureRandom.nextBoolean();

For generating int values we can pass an upper bound as a parameter:

int randomInt = secureRandom.nextInt(upperBound);

In addition, we can generate a stream of values for int, double and long:

IntStream randomIntStream = secureRandom.ints();
LongStream randomLongStream = secureRandom.longs();
DoubleStream randomDoubleStream = secureRandom.doubles();

For all streams we can explicitly set the stream size:

IntStream intStream = secureRandom.ints(streamSize);

This class offers several other options ( eg, choosing the underlying random number generator) that are out of the scope of this question.

Answer 66

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)
        {
            System.out.println(cd-abc);
            break;
        }
    }

Answer 67

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)
        numbers.sort(null);

    return numbers;
}

It generates numbers from 1-Randombound in this example.

Answer 68

A simple way to generate n random numbers between a and b eg 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 .

Answer 69

public static void main(String[] args) {

    Random ran = new Random();

    int min, max;
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter min range:");
    min = sc.nextInt();
    System.out.println("Enter max range:");
    max = sc.nextInt();
    int num = ran.nextInt(min);
    int num1 = ran.nextInt(max);
    System.out.println("Random Number between given range is " + num1);

}

Answer 70

Here's a function that returns exactly one integer random number in a range defined by lowerBoundIncluded and upperBoundIncluded , as requested by user42155

SplittableRandom splittableRandom = new SplittableRandom();

BiFunction<Integer,Integer,Integer> randomInt = (lowerBoundIncluded, upperBoundIncluded)
    -> splittableRandom.nextInt(lowerBoundIncluded, upperBoundIncluded + 1);

randomInt.apply( …, … ); // gets the random number

…or shorter for the one-time generation of a random number

new SplittableRandom().nextInt(lowerBoundIncluded, upperBoundIncluded + 1);

Answer 71

There are many ways to generate random integers within a specific range in java.We'll explore different ways of generating random numbers within a range.

Math.Random

Let's use the Math.random method to generate a random number in a given range [min, max).For generating random numbers within a range using Math.random(), follow the steps below:

• Declare the minimum value of the range

• Declare the maximum value of the range

• Use the formula Math.floor(Math.random()*(max-min+1)+min) to generate values with the min and the max value inclusive.

   public class MathRandomExample { public static void main(String args[]) { int min = 1; int max = 10; System.out.println("random value in int from "+min+" to "+max+ ":"); int random_int = (int)Math.floor(Math.random()*(max-min+1)+min); System.out.println(random_int); } }

  Note: This method can only be used if you need an integer or float random value.

Random Class

We can also use an instance of java.util.Random to do the same.

Let's make use of the java.util.Random.nextInt method to get a random number.

public class RandomNextIntExample {
public static void main(String args[]) {
    int min = 1;
    int max = 10;

    Random random = new Random();
    System.out.println(random.nextInt(max - min) + min);

   }
}

The java.util.Random.ints method returns an IntStream of random integers. So, we can utilize the java.util.Random.ints method and return a random number:

public class RandomIntsExample {
public static void main(String args[]) {
    int min = 1;
    int max = 10;

    Random random = new Random();
    System.out.println(random.ints(min, max)
        .findFirst()
        .getAsInt());

    }
}

Answer 72

How to generate random number in java

---

There are many ways in java to generate random numbers. But I will show only 2 ways.

1. Using the random() Method
2. Using the Random Class

Using the Math.random() Method

We can use the following formula if we want to a generate random number between a specified range.

Math.random() * (max - min + 1) + min  

Integer generation:

int b = (int)(Math.random()*(max-min+1)+min);  
System.out.println(b);  

Double generation:

double a = Math.random()*(max-min+1)+min;   
System.out.println(a);  

Using the Random Class

Another way to generate a random number is to use the Java Random class of the java.util package.

First, import the class java.util.Random .

import java.util.Random;

Create an object of the Random class.

Random random = new Random();   

Invoke any of the following methods:

1. nextInt(min, max),
2. nextDouble(min, max)
3. etc

Random Integer Generation:

// Generates random integers 0 to 49  
int x = random.nextInt(50);   
// Generates random integers 100 to 1000
int y = random.nextInt(100, ,1001);   
// Prints random integer values  

Random Double Generation:

// Generates Random doubles values
double a = random.nextDouble();   
// Generates Random doubles values 100 to 1000
double b = random.nextDouble(100, 1001);   

Answer 73

We can use Random class to generate value in between specified range by passing origin and bound value. Note that bound value is exclusive.

new Random().nextInt(1, 10)

Answer 74

If you don't want to re-invent the wheel, her is a one liner, no-brainer solution:

RandomStringUtils.randomNumeric(count); // where count is the number of digits you want in the random number.

Apache Utils provides many options to generate whatever format of random number you wish for.

Answer 75

Because the Android question redirects here, this is how you would do it with Kotlin:

val r = (0..10).random() // A random integer between 0 and 10 inclusive

This works with Kotlin 1.3 and higher. Refer to this answer .

Answer 76

If we have security-sensitive apps, we should use SecureRandom.

    import java.security.SecureRandom;

    public class Main {
      public static void main(String... args){
        int max=13;
        int min =1;
        int randomWithSecureRandom = new SecureRandom().nextInt(max - min) + min;
        System.out.println(randomWithSecureRandom);
      }
    }

Answer 77

The following snippet will give a random value between 0 and 10000:

import java.util.Random;
public class Main
{
    public static void main(String[] args) {
        Random rand = new Random();
        System.out.printf("%04d%n", rand.nextInt(10000));
    }
}

Answer 78

You can do as below.

import java.util.Random;
public class RandomTestClass {

    public static void main(String[] args) {
        Random r = new Random();
        int max, min;
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter maximum value : ");
        max = scanner.nextInt();
        System.out.println("Enter minimum value : ");
        min = scanner.nextInt();
        int randomNum;
        randomNum = r.nextInt(max) + min;
        System.out.println("Random Number : " + randomNum);
    }

}