CODEWITHSUNDEEP
javaoptimizationjava-8java-stream
Ahmad Al-Khazraji
Ahmad Al-Khazraji

Reputation: 127

Java 8: check for common elements in two lists using streams

I'm looking for a statment to check if there is any match in two lists of Users, according to Username.

List<User> a;
List<User> b;
for (User user : a) {
    for (User newUser : b) {
        if (user.getName().equals(newUser.getName())) {
        }
    }
}

How can I write this in java 8? Somthing like this:

List<User> intersect = a.stream()
                    .filter(User::getName)
                    .collect(Collectors.toList());

Upvotes: 2

Views: 9260

Answers (3)

Joop Eggen
Joop Eggen

Reputation: 109547

When User is correctly defined with a hashCode and equals (otherwise you might try TreeSet instead of HashSet), do set-operations:

Set<User> common = new HashSet<>(a);
common.retainAll(b);

If User.getName is not used for equality:

Set<User> common = new TreeSet<>(Comparator.comparing(User::getName));
common.addAll(a);
common.retainAll(b);

Two nested for loops on lists (also as streams) would have complexity O(N²), whereas this is O(N.log N).

Upvotes: 3

Naman
Naman

Reputation: 31878

One way to do that using Stream.anyMatch(this would break within if) could be :

a.stream().filter(user -> b.stream().anyMatch(newUser -> user.getName().equals(newUser.getName())))
          .map(User::getName)
          .forEach(System.out::println); // logic inside 'if' here (print for e.g.)

If you want to repeat the loop(if logic) for all such matches :

a.forEach(user -> b.stream()
                   .filter(newUser -> user.getName().equals(newUser.getName()))
                   .map(newUser -> user.getName())
                   .forEach(System.out::println));

Upvotes: 1

Sukhpal Singh
Sukhpal Singh

Reputation: 2280

You can do something like below:

List<User> intersect = a.stream()
                     .filter(b::contains)
                     .collect(Collectors.toList());

You need to override equals and hashCode methods in User. For optimization, you can convert b to HashSet first.

Upvotes: 2

Related Questions