我已经习惯了在Objective c中使用数据字典来实现这一目标，而在Java for Android中却很难获得类似的结果。我最终创建了一个自定义类，然后只做了一个自定义类的hashmap。

public class Test1 {
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.addview);

//create the datastring
    HashMap<Integer, myClass> hm = new HashMap<Integer, myClass>();
    hm.put(1, new myClass("Car", "Small", 3000));
    hm.put(2, new myClass("Truck", "Large", 4000));
    hm.put(3, new myClass("Motorcycle", "Small", 1000));

//pull the datastring back for a specific item.
//also can edit the data using the set methods.  this just shows getting it for display.
    myClass test1 = hm.get(1);
    String testitem = test1.getItem();
    int testprice = test1.getPrice();
    Log.i("Class Info Example",testitem+Integer.toString(testprice));
}
}

//custom class.  You could make it public to use on several activities, or just include in the activity if using only here
class myClass{
    private String item;
    private String type;
    private int price;

    public myClass(String itm, String ty, int pr){
        this.item = itm;
        this.price = pr;
        this.type = ty;
    }

    public String getItem() {
        return item;
    }

    public void setItem(String item) {
        this.item = item;
    }

    public String getType() {
        return item;
    }

    public void setType(String type) {
        this.type = type;
    }

    public int getPrice() {
        return price;
    }

    public void setPrice(int price) {
        this.price = price;
    }

}

HashMap<Integer,ArrayList<String>> map = new    HashMap<Integer,ArrayList<String>>();

ArrayList<String> list = new ArrayList<String>();
list.add("abc");
list.add("xyz");
map.put(100,list);

是也不是。解决方案是为包含与键对应的2(3或更多)值的值构建一个Wrapper类。

只是为了记录，纯JDK8解决方案将使用Map::compute方法:

map.compute(key, (s, strings) -> strings == null ? new ArrayList<>() : strings).add(value);

如

public static void main(String[] args) {
    Map<String, List<String>> map = new HashMap<>();

    put(map, "first", "hello");
    put(map, "first", "foo");
    put(map, "bar", "foo");
    put(map, "first", "hello");

    map.forEach((s, strings) -> {
        System.out.print(s + ": ");
        System.out.println(strings.stream().collect(Collectors.joining(", ")));
    });
}

private static <KEY, VALUE> void put(Map<KEY, List<VALUE>> map, KEY key, VALUE value) {
    map.compute(key, (s, strings) -> strings == null ? new ArrayList<>() : strings).add(value);
}

输出:

bar: foo
first: hello, foo, hello

注意，为了确保在多个线程访问此数据结构时的一致性，需要使用ConcurrentHashMap和CopyOnWriteArrayList作为实例。

你可以:

使用具有列表作为值的映射。地图< KeyType、列表< ValueType > >。 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 >。 使用类似类的元组(节省了创建大量包装器)。映射<KeyType，元组<Value1Type, Value2Type>>。 同时使用多个地图。

例子

1. 以list为值的映射

// 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没有绑定到恰好两个值。

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
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;

在我看来，这是最好的解决办法。

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"];

这种解决方案的缺点是，两个映射之间的关联并不明显，编程错误可能会导致两个映射不同步。

String key= "services_servicename"

ArrayList<String> data;

for(int i = 0; i lessthen data.size(); i++) {
    HashMap<String, String> servicesNameHashmap = new HashMap<String, String>();
    servicesNameHashmap.put(key,data.get(i).getServiceName());
    mServiceNameArray.add(i,servicesNameHashmap);
}

我得到了最好的成绩。

你只需要创建新的HashMap

HashMap<String, String> servicesNameHashmap = new HashMap<String, String>();

在for循环中。它将具有相同的效果，如相同的键和多个值。