如何使用異步和使用多線程在Java中獲取列表列表的交集

Question

我有一個這樣的Person類：

public class Person {
    public String name;
    public List<String> hobbies;


    public Person(String name, List<String> hobbies) {
        this.name = name;
        this.hobbies = hobbies;
    } 
    //setter and getter
}

現在假設我有一個Person對象列表，我想異步獲取愛好的交集。 我知道如何正常獲取兩個列表的交集，但我想以異步和多線程方式執行此操作。 一世

Answer 1

你可以這樣做：

List<Person> persons = new ArrayList<>();
persons.add(new Person("hatef", Arrays.asList("a", "b", "c", "t")));
persons.add(new Person("hatef", Arrays.asList("a", "b", "t")));
persons.add(new Person("hatef", Arrays.asList("t", "a", "b")));
persons.add(new Person("hatef", Arrays.asList("a", "b", "f", "t")));
persons.add(new Person("hatef", Arrays.asList("a", "t", "b", "z")));

然后使用CompletableFuture

Map<String, Integer> all = new ConcurrentHashMap<>();
ExecutorService executorService = Executors.newFixedThreadPool(Math.max(10, persons.size()));
persons
        .stream()
        .map(person -> CompletableFuture.supplyAsync(() -> person, executorService))
        .map(p -> p.thenApply(Person::getHobbies))
        .map(p -> p.thenAccept(hobbies -> hobbies.forEach(hobby -> all.compute(hobby, (key, value) -> value == null ? 1 : ++value))))
        .map(CompletableFuture::join)
        .forEach(x -> {}); // just dummy consumer to make stream triggers

int n = persons.size();
Set<String> result = all
            .entrySet()
            .stream()
            .filter(entry -> entry.getValue() == n)
            .map(Map.Entry::getKey)
            .collect(Collectors.toSet());
System.out.println(result);

這導致：

[a, b, t]

如果您在愛好列表中有訂單，我認為您可以使用最長公共前綴問題的想法並且它是有效的

我同意@JoachimSauer 的評論