Java – Generate random integers in a range
In this article, we will show you three ways to generate random integers in a range.
- java.util.Random.nextInt
- Math.random
- 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).
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).
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
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).
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
Really Thank You, You Saved My Day ?
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!
How to generate dynamic regex for numeric range in java
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()
I believe the reason is because of time complexity. If my theory is correct, it takes more time to call “findFirst” in a stream of “Long.MAX_VALUE” numbers than to call “limit(1).findFirst()” because you’ll be operating in a stream of only one number. And, according to IntStream Javadoc, limit() is a cheap operation. That tells me that the overall time complexity of both calls must be less than the time complexity of calling findFirst straight up.
i dont know. Dont ask me.
Thank you, I have just been searching for info about this subject for a long time and yours is the greatest I’ve found out so far. However, what about the bottom line? Are you positive concerning the supply? ceefcffkaakbbkkg
Thanks, Worked to me. Looks precise.
1.3 code snippet (max – min) + 1 will not work if max parameter is Integer.MAX_VALUE
how to make Integer.MAX_VALUE inclusive? Impossible?
static int getRandomNumberInRange(int min, int max) {
if (min >= max) {
throw new IllegalArgumentException(“max must be greater than min”);
}
Random random = new Random();
int bound = (max == Integer.MAX_VALUE) ? (max – min) : (max – min) + 1;
return random.nextInt(bound) + min;
// return random.ints(min, bound).limit(1).findFirst().orElse(min);
}
I cant make Integer.MAX_VALUE inclusive, but at least it will work with Integer.MAX_VALUE as max paramenter exclusive.
How can I generate a number in range of [-100 ,100] I really stuck.
The only thing I can do is this but that’s not what I need
Random random = new Random();
int rand = random.nextInt(100);
int rand1 =(-100 + random.nextInt(100));
Is there an explanation of why the first one works? I get how it works but why it does 😀
How can I save the generated numbers as integers (variables)? The console says “intstreams can’t be converted to int”
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. 😀
* @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 and 99
Random r = new Random();
for(int i = 0; i < 10; i++)
System.out.println(r.nextInt(98 + 1)+ 1);
// (99max) + (1min) + 1min
}
}