Java流-从另外两个列表中获得一个“对称差异列表”

Question

public void testDisjointLists() {
    List<Car> bigCarList = get5DefaultCars();
    List<Car> smallCarList = get3DefaultCars();

    //Get cars that exists in both lists
    List<Car> intersect = bigCarList.stream().filter(smallCarList::contains).collect(Collectors.toList());

    //Get all cars in both list as one list
    List<Car> union = Stream.concat(bigCarList.stream(), smallCarList.stream()).distinct().collect(Collectors.toList());

    //Get all cars that only exist in one list
    //List<Car> disjoint = ???

}

public List<Car> get5DefaultCars() {
    List<Car> cars = get3DefaultCars();
    cars.add(new Car("Toyota Corolla", 2008));
    cars.add(new Car("Ford Focus", 2010));
    return cars;
}

public List<Car> get3DefaultCars() {
    List<Car> cars = new ArrayList<>();
    cars.add(new Car("Volvo V70", 1990));
    cars.add(new Car("BMW I3", 1999));
    cars.add(new Car("Audi A3", 2005));
    return cars;
}

class Car {
    private int releaseYear;
    private String name;
    public Car(String name) {
        this.name = name;
    }
    public Car(String name, int releaseYear) {
        this.name = name;
        this.releaseYear = releaseYear;
    }

    //Overridden equals() and hashCode()
}

Answer 1

基于您自己的代码，有一个简单的解决方案：

List<Car> disjoint = Stream.concat(
    bigCarList.stream().filter(c->!smallCarList.contains(c)),
    smallCarList.stream().filter(c->!bigCarList.contains(c))
).collect(Collectors.toList());

只需筛选一个列表中不包含的所有项，反之亦然，并将两个结果连接起来。这对于小列表非常有效，在考虑优化的解决方案（如散列或使结果distinct()）之前，您应该问问自己，如果既不想要列表，也不想要重复列表，也不想要特定的顺序，为什么要使用列表。

似乎您实际上想要集而不是列表。如果使用sets，Tagir Valeev的解决方案是合适的。但是它不符合lists的实际语义，也就是说，如果源列表包含重复项，它就不起作用。

Set<Car> disjoint = Stream.concat(bigCarSet.stream(), smallCarSet.stream())
  .collect(Collectors.toMap(Function.identity(), t->true, (a,b)->null))
  .keySet();