你可以：

1. 使用具有列表作为值的地图。Map<KeyType, List<ValueType>>。
2. 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>。
3. 使用类似类的元组（节省创建许多包装器）。Map<KeyType, Tuple<Value1Type, Value2Type>>。
4. 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，你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。