你可以：
- 使用具有列表作为值的地图。
Map<KeyType, List<ValueType>>。
- 创建一个新的包装器类，并将该包装器的实例放置在地图中。
Map<KeyType, WrapperType>。
- 使用类似类的元组（节省创建许多包装器）。
Map<KeyType, Tuple<Value1Type, Value2Type>>。
- 并排使用多个地图。
例子
1.使用列表作为值进行映射
// create our map
Map<String, List<Person>> peopleByForename = new HashMap<>();
// populate it
List<Person> people = new ArrayList<>();
people.add(new Person("Bob Smith"));
people.add(new Person("Bob Jones"));
peopleByForename.put("Bob", people);
// read from it
List<Person> bobs = peopleByForename["Bob"];
Person bob1 = bobs[0];
Person bob2 = bobs[1];
这种方法的缺点是列表未完全绑定到两个值。
2.使用包装器类
// define our wrapper
class Wrapper {
public Wrapper(Person person1, Person person2) {
this.person1 = person1;
this.person2 = person2;
}
public Person getPerson1 { return this.person1; }
public Person getPerson2 { return this.person2; }
private Person person1;
private Person person2;
}
// create our map
Map<String, Wrapper> peopleByForename = new HashMap<>();
// populate it
Wrapper people = new Wrapper();
peopleByForename.put("Bob", new Wrapper(new Person("Bob Smith"),
new Person("Bob Jones"));
// read from it
Wrapper bobs = peopleByForename.get("Bob");
Person bob1 = bobs.getPerson1;
Person bob2 = bobs.getPerson2;
这种方法的缺点是您必须为所有这些非常简单的容器类编写很多样板代码。
3.使用元组
// you'll have to write or download a Tuple class in Java, (.NET ships with one)
// create our map
Map<String, Tuple2<Person, Person> peopleByForename = new HashMap<>();
// populate it
peopleByForename.put("Bob", new Tuple2(new Person("Bob Smith",
new Person("Bob Jones"));
// read from it
Tuple<Person, Person> bobs = peopleByForename["Bob"];
Person bob1 = bobs.Item1;
Person bob2 = bobs.Item2;
我认为这是最好的解决方案。
4.多个地图
// create our maps
Map<String, Person> firstPersonByForename = new HashMap<>();
Map<String, Person> secondPersonByForename = new HashMap<>();
// populate them
firstPersonByForename.put("Bob", new Person("Bob Smith"));
secondPersonByForename.put("Bob", new Person("Bob Jones"));
// read from them
Person bob1 = firstPersonByForename["Bob"];
Person bob2 = secondPersonByForename["Bob"];
此解决方案的缺点是，两个映射之间的关联并不明显，程序错误可能会导致两个映射不同步。