# Java – Generate random integers in a range

In this article, we will show you three ways to generate random integers in a range.

1. java.util.Random.nextInt
2. Math.random
3. java.util.Random.ints (Java 8)

## 1. java.util.Random

This `Random().nextInt(int bound)` generates a random integer from 0 (inclusive) to bound (exclusive).

1.1 Code snippet. For `getRandomNumberInRange(5, 10)`, this will generates a random integer between 5 (inclusive) and 10 (inclusive).

``````
private static int getRandomNumberInRange(int min, int max) {

if (min >= max) {
throw new IllegalArgumentException("max must be greater than min");
}

Random r = new Random();
return r.nextInt((max - min) + 1) + min;
}
``````

1.2 What is (max – min) + 1) + min?

Above formula will generates a random integer in a range between min (inclusive) and max (inclusive).

``````
//Random().nextInt(int bound) = Random integer from 0 (inclusive) to bound (exclusive)

//1. nextInt(range) = nextInt(max - min)
new Random().nextInt(5);  // [0...4] [min = 0, max = 4]
new Random().nextInt(6);  // [0...5]
new Random().nextInt(7);  // [0...6]
new Random().nextInt(8);  // [0...7]
new Random().nextInt(9);  // [0...8]
new Random().nextInt(10); // [0...9]
new Random().nextInt(11); // [0...10]

//2. To include the last value (max value) = (range + 1)
new Random().nextInt(5 + 1)  // [0...5] [min = 0, max = 5]
new Random().nextInt(6 + 1)  // [0...6]
new Random().nextInt(7 + 1)  // [0...7]
new Random().nextInt(8 + 1)  // [0...8]
new Random().nextInt(9 + 1)  // [0...9]
new Random().nextInt(10 + 1) // [0...10]
new Random().nextInt(11 + 1) // [0...11]

//3. To define a start value (min value) in a range,
//   For example, the range should start from 10 = (range + 1) + min
new Random().nextInt(5 + 1)  + 10 // [0...5]  + 10 = [10...15]
new Random().nextInt(6 + 1)  + 10 // [0...6]  + 10 = [10...16]
new Random().nextInt(7 + 1)  + 10 // [0...7]  + 10 = [10...17]
new Random().nextInt(8 + 1)  + 10 // [0...8]  + 10 = [10...18]
new Random().nextInt(9 + 1)  + 10 // [0...9]  + 10 = [10...19]
new Random().nextInt(10 + 1) + 10 // [0...10] + 10 = [10...20]
new Random().nextInt(11 + 1) + 10 // [0...11] + 10 = [10...21]

// Range = (max - min)
// So, the final formula is ((max - min) + 1) + min

//4. Test [10...30]
// min = 10 , max = 30, range = (max - min)
new Random().nextInt((max - min) + 1) + min
new Random().nextInt((30 - 10) + 1) + 10
new Random().nextInt((20) + 1) + 10
new Random().nextInt(21) + 10    //[0...20] + 10 = [10...30]

//5. Test [15...99]
// min = 15 , max = 99, range = (max - min)
new Random().nextInt((max - min) + 1) + min
new Random().nextInt((99 - 15) + 1) + 15
new Random().nextInt((84) + 1) + 15
new Random().nextInt(85) + 15    //[0...84] + 15 = [15...99]

//Done, understand?
``````

1.3 Full examples to generate 10 random integers in a range between 5 (inclusive) and 10 (inclusive).

TestRandom.java
``````
package com.mkyong.example.test;

import java.util.Random;

public class TestRandom {

public static void main(String[] args) {

for (int i = 0; i < 10; i++) {
System.out.println(getRandomNumberInRange(5, 10));
}

}

private static int getRandomNumberInRange(int min, int max) {

if (min >= max) {
throw new IllegalArgumentException("max must be greater than min");
}

Random r = new Random();
return r.nextInt((max - min) + 1) + min;
}

}
``````

Output.

``````
7
6
10
8
9
5
7
10
8
5
``````

## 2. Math.random

This `Math.random()` gives a random double from 0.0 (inclusive) to 1.0 (exclusive).

2.1 Code snippet. Refer to 1.2, more or less it is the same formula.

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

2.2 Full examples to generate 10 random integers in a range between 16 (inclusive) and 20 (inclusive).

TestRandom.java
``````
package com.mkyong.example.test;

public class TestRandom {

public static void main(String[] args) {

for (int i = 0; i < 10; i++) {
System.out.println(getRandomNumberInRange(16, 20));
}

}

private static int getRandomNumberInRange(int min, int max) {

if (min >= max) {
throw new IllegalArgumentException("max must be greater than min");
}

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

}
``````

Output.

``````
17
16
20
19
20
20
20
17
20
16
``````
Note
The `Random.nextInt(n)` is more efficient than `Math.random() * n`, read this Oracle forum post.

## 3. Java 8 Random.ints

In Java 8, new methods are added in `java.util.Random`

``````
public IntStream ints(int randomNumberOrigin, int randomNumberBound)
public IntStream ints(long streamSize, int randomNumberOrigin, int randomNumberBound)
``````

This `Random.ints(int origin, int bound)` or `Random.ints(int min, int max)` generates a random integer from origin (inclusive) to bound (exclusive).

3.1 Code snippet.

``````
private static int getRandomNumberInRange(int min, int max) {

Random r = new Random();
return r.ints(min, (max + 1)).findFirst().getAsInt();

}
``````

3.2 Full examples to generate 10 random integers in a range between 33 (inclusive) and 38 (inclusive).

TestRandom.java
``````
package com.mkyong.form.test;

import java.util.Random;

public class TestRandom {

public static void main(String[] args) {

for (int i = 0; i < 10; i++) {
System.out.println(getRandomNumberInRange(33, 38));
}

}

private static int getRandomNumberInRange(int min, int max) {

Random r = new Random();
return r.ints(min, (max + 1)).limit(1).findFirst().getAsInt();

}

}
``````

Output.

``````
34
35
37
33
38
37
34
35
36
37
``````

3.3 Extra, for self-reference.

Generates random integers in a range between 33 (inclusive) and 38 (exclusive), with stream size of 10. And print out the items with `forEach`.

``````
//Java 8 only
new Random().ints(10, 33, 38).forEach(System.out::println);
``````

Output.

``````
34
37
37
34
34
35
36
33
37
34
``````

## References

#### mkyong

Founder of Mkyong.com, love Java and open source stuff. Follow him on Twitter. If you like my tutorials, consider make a donation to these charities. Read all published posts by

Guest
Koussay

Really Thank You, You Saved My Day ?

Guest
shiva

How to generate dynamic regex for numeric range in java

Guest
Smithf203

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Guest
Aisht

Why do we use limit(1) in below snippet? Using only findFirst() should return single random integer, isn’t it?
r.ints(min, (max + 1)).limit(1).findFirst().getAsInt()

Guest
morrris

i dont know. Dont ask me.

Guest
gyf

I find a little error,as
1.3 Full examples to generate 10 random integers in a range between 5 (inclusive) and 10 (inclusive).
java doc is
nextInt(int bound)
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive).
Thank you all the same!

Guest
Fernando Hdez

How can I save the generated numbers as integers (variables)? The console says “intstreams can’t be converted to int”

Guest
Wroh

Is there an explanation of why the first one works? I get how it works but why it does ðŸ˜€

Guest
BPC

This code is the easiest way to return 10 random numbers between 1 and 99. I took all your ideas and came up with this brief but effective code. Just change the values of 99,1,1 to your min and max to get your #s. If you use 99 as your max, randomly 99 + 1 will make the code generate 100, so if you really want max of 99, use 98 in this code. ðŸ˜€ * @author balcopc */ import java.util.Random; public class RandomNumberProj { public static void main(String[] args) { System.out.println(“Random Numbers: “); //print ten random numbers between 1… Read more »

Guest
sai kiran chary

Thanks, Worked to me. Looks precise.