Generate a List of sequential integers in Java

This post will discuss how to generate a list of sequential integers in Java. The solution should create a list with ranges of numbers between 1 and n.

1. Using `IntStream.range()` method

In Java 8 or later, this can be easily done using Streams without looping or using third-party libraries. The idea is to use the IntStream.range(…) method to generate a stream of increasing integers between the specified indices. To get a list of integers, you can box the primitive int stream and collect the stream elements into a list.

import java.util.List;

import java.util.stream.Collectors;

import java.util.stream.IntStream;

public class Main

{

public static void main(String[] args) {

int n = 5;

List<Integer> list = IntStream.range(1, n + 1)

.boxed()

.collect(Collectors.toList());

System.out.println(list);

}

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Output:

[1, 2, 3, 4, 5]

2. Using `IntStream.iterate()` method

Another alternative is to use the IntStream.iterate(…) method to get an infinite sequential ordered IntStream. It is produced by applying a function f to an initial element x, thereby producing a Stream consisting of x, f(x), f(f(x)), etc.

Here’s complete usage of this method:

import java.util.List;

import java.util.stream.Collectors;

import java.util.stream.IntStream;

public class Main

{

public static void main(String[] args) {

int n = 5;

List<Integer> list = IntStream.iterate(1, i -> i + 1)

.limit(n)

.boxed()

.collect(Collectors.toList());

System.out.println(list);

}

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Output:

[1, 2, 3, 4, 5]

The above solution restricts the number of elements in the infinite stream to n using the limit() method. A better way to do it in Java 9 and above is with the overloaded 3-arg IntStream.iterate(…) method, which returns a finite sequential ordered IntStream, terminating on satisfying the specified predicate.

import java.util.List;

import java.util.stream.Collectors;

import java.util.stream.IntStream;

public class Main

{

public static void main(String[] args) {

int n = 5;

List<Integer> list = IntStream.iterate(1, i -> i <= n, i -> i + 1)

.boxed()

.collect(Collectors.toList());

System.out.println(list);

}

Download Run Code

Output:

[1, 2, 3, 4, 5]

3. Using Guava Ranges

Another plausible solution is to use Guava Ranges to create a sorted set of contiguous values (ContiguousSet) with the elements of a range of a discrete domain. To get a list, call the asList() method on the sorted set.

Its usage is demonstrated below.

import com.google.common.collect.ContiguousSet;

import com.google.common.collect.DiscreteDomain;

import com.google.common.collect.Range;

import java.util.List;

public class Main

{

public static void main(String[] args) {

int n = 5;

List<Integer> list = ContiguousSet.create(Range.closed(1, n),

DiscreteDomain.integers()).asList();

System.out.println(list);

}

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Output:

[1, 2, 3, 4, 5]

That’s all about generating a list of sequential integers in Java.

Generate a List of sequential integers in Java

1. Using IntStream.range() method

2. Using IntStream.iterate() method

3. Using Guava Ranges

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1. Using `IntStream.range()` method

2. Using `IntStream.iterate()` method