如果我将相同的键多次传递给HashMap的put方法，原始值会发生什么变化?如果值重复呢?我没找到任何关于这个的文件。
情况1:键的覆盖值
Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","not one");
mymap.put("1","surely not one");
System.out.println(mymap.get("1"));
我们得到的肯定不是1。
案例2:重复值
Map mymap = new HashMap();
mymap.put("1","one");
mymap.put("1","not one");
mymap.put("1","surely not one");
// The following line was added:
mymap.put("1","one");
System.out.println(mymap.get("1"));
我们得到一个。
但是其他的值会怎样呢?我在给一个学生教授基础知识，有人问我这个问题。Map是否像一个引用最后一个值的桶(但在内存中)?
我总是用:
HashMap<String, ArrayList<String>> hashy = new HashMap<String, ArrayList<String>>();
如果我想把多个东西应用到一个标识键上。
public void MultiHash(){
HashMap<String, ArrayList<String>> hashy = new HashMap<String, ArrayList<String>>();
String key = "Your key";
ArrayList<String> yourarraylist = hashy.get(key);
for(String valuessaved2key : yourarraylist){
System.out.println(valuessaved2key);
}
}
你可以做一些这样的事情，为自己创建一个迷宫!
public void LOOK_AT_ALL_THESE_HASHMAPS(){
HashMap<String, HashMap<String, HashMap<String, HashMap<String, String>>>> theultimatehashmap = new HashMap <String, HashMap<String, HashMap<String, HashMap<String, String>>>>();
String ballsdeep_into_the_hashmap = theultimatehashmap.get("firststring").get("secondstring").get("thirdstring").get("forthstring");
}
你可以在map# put(K, V)的javadoc中找到答案(它实际上会返回一些东西):
public V put(K key,
V value)
Associates the specified value with the specified key in this map
(optional operation). If the map
previously contained a mapping for
this key, the old value is replaced by
the specified value. (A map m is said
to contain a mapping for a key k if
and only if m.containsKey(k) would
return true.)
Parameters:
key - key with which the specified value is to be associated.
value - value to be associated with the specified key.
Returns:
previous value associated with specified key, or null if there was no
mapping for key. (A null return can also indicate that the map previously associated null with the specified key, if the implementation supports null values.)
如果你在调用mymap时不分配返回值。Put ("1"， "a string")，它就变成不被引用的，因此有资格进行垃圾收集。
我总是用:
HashMap<String, ArrayList<String>> hashy = new HashMap<String, ArrayList<String>>();
如果我想把多个东西应用到一个标识键上。
public void MultiHash(){
HashMap<String, ArrayList<String>> hashy = new HashMap<String, ArrayList<String>>();
String key = "Your key";
ArrayList<String> yourarraylist = hashy.get(key);
for(String valuessaved2key : yourarraylist){
System.out.println(valuessaved2key);
}
}
你可以做一些这样的事情，为自己创建一个迷宫!
public void LOOK_AT_ALL_THESE_HASHMAPS(){
HashMap<String, HashMap<String, HashMap<String, HashMap<String, String>>>> theultimatehashmap = new HashMap <String, HashMap<String, HashMap<String, HashMap<String, String>>>>();
String ballsdeep_into_the_hashmap = theultimatehashmap.get("firststring").get("secondstring").get("thirdstring").get("forthstring");
}
它是键/值特性，你不能有多个值的重复键，因为当你想获得实际值时，哪个值属于输入的键在你的例子中，当你想获得值“1”时，它是哪一个?!这就是为什么每个值都有唯一的键，但你可以通过Java标准库有一个技巧:
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
public class DuplicateMap<K, V> {
private Map<K, ArrayList<V>> m = new HashMap<>();
public void put(K k, V v) {
if (m.containsKey(k)) {
m.get(k).add(v);
} else {
ArrayList<V> arr = new ArrayList<>();
arr.add(v);
m.put(k, arr);
}
}
public ArrayList<V> get(K k) {
return m.get(k);
}
public V get(K k, int index) {
return m.get(k).size()-1 < index ? null : m.get(k).get(index);
}
}
你可以这样使用它:
public static void main(String[] args) {
DuplicateMap<String,String> dm=new DuplicateMap<>();
dm.put("1", "one");
dm.put("1", "not one");
dm.put("1", "surely not one");
System.out.println(dm.get("1"));
System.out.println(dm.get("1",1));
System.out.println(dm.get("1", 5));
}
打印结果为:
[one, not one, surely not one]
not one
null
