CODEWITHSUNDEEP
javaarraysperformance
Joel
Joel

Reputation: 1843

Fastest way to get the first n elements of a List into an Array

What is the fastest way to get the first n elements of a list stored in an array?

Considering this as the scenario:

int n = 10;
ArrayList<String> in = new ArrayList<>();
for(int i = 0; i < (n+10); i++)
  in.add("foobar");

Option 1:

String[] out = new String[n];
for(int i = 0; i< n; i++)
    out[i]=in.get(i);

Option 2:

String[] out = (String[]) (in.subList(0, n)).toArray();

Option 3: Is there a faster way? Maybe with Java8-streams?

Upvotes: 78

Views: 240845

Answers (7)

Sai Nikhil A
Sai Nikhil A

Reputation: 1

arrayList.stream().limit(n).toArray();

n = maxSize in length

This will help you to get the max size of the Array that you require.

Upvotes: 0

FarVoyager
FarVoyager

Reputation: 59

Use .take(n) operator on your list

Upvotes: 0

Jack Cole
Jack Cole

Reputation: 1804

Option 3

Iterators are faster than using the get operation, since the get operation has to start from the beginning if it has to do some traversal. It probably wouldn't make a difference in an ArrayList, but other data structures could see a noticeable speed difference. This is also compatible with things that aren't lists, like sets.

String[] out = new String[n];
Iterator<String> iterator = in.iterator();
for (int i = 0; i < n && iterator.hasNext(); i++)
    out[i] = iterator.next();

Upvotes: 1

user2280949
user2280949

Reputation: 77

Use: Arrays.copyOf(yourArray,n);

Upvotes: -6

src3369
src3369

Reputation: 2019

Assumption:

list - List<String>

Using Java 8 Streams,

  • to get first N elements from a list into a list,

    List<String> firstNElementsList = list.stream().limit(n).collect(Collectors.toList());

  • to get first N elements from a list into an Array,

    String[] firstNElementsArray = list.stream().limit(n).collect(Collectors.toList()).toArray(new String[n]);

Upvotes: 157

ZhongYu
ZhongYu

Reputation: 19682

It mostly depends on how big n is.

If n==0, nothing beats option#1 :)

If n is very large, toArray(new String[n]) is faster.

Upvotes: 3

Elliott Frisch
Elliott Frisch

Reputation: 201487

Option 1 Faster Than Option 2

Because Option 2 creates a new List reference, and then creates an n element array from the List (option 1 perfectly sizes the output array). However, first you need to fix the off by one bug. Use < (not <=). Like,

String[] out = new String[n];
for(int i = 0; i < n; i++) {
    out[i] = in.get(i);
}

Upvotes: 12

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