Java HashMap在同一个键下有多个值

您可以:

1. 使用具有列表作为值的映射.Map<KeyType, List<ValueType>>
2. 创建一个新的包装器类，并在映射中放置该包装器的实例.Map<KeyType, WrapperType>
3. 使用类似类的元组(省go 了创建大量包装器的麻烦).Map<KeyType, Tuple<Value1Type, Value2Type>>美元.
4. 并排使用多张 map .

示例

1. Map with list as the value

// 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，列表不能精确绑定到两个值.

2. Using wrapper class

// 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
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，必须为所有这些非常简单的容器类编写大量的boiler plate代码.

3. Using a tuple个

// 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;

在我看来，这是最好的解决办法.

4. Multiple maps

// 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"];

这种解决方案的缺点是，这两个映射并不明显相关，编程错误可能会导致这两个映射不同步.