如何存储具有相同密钥的散列表中的多个对象？ [英] How to store multiple objects from a hashmap that has the same key?

问题描述

编辑

我试过这个>具有多个值的HashMap在同一个键下，我的hashMap现在看起来像这样 HashMap< String，List< Place>> ; placeMap = new HashMap<>（）;

还试图把Object放在Place上（放置是我的超类）。但是，当我现在创建我的子类并想将它们添加到HashMap时，我会得到：

方法put（String，List）类型 HashMap< String，List< Place>> 不适用于参数（String，NamedPlace）

和

方法put（String，List）的类型为 HashMap< String，List< Place>> 不适用于参数（字符串，DescPlace）是我的添加，它创建了错误：

pre> NamedPlace p = new NamedPlace（x，y，answer，col，cat）;
placeMap.put（answer，p）;


---

  DescPlace dp = new DescPlace（x，y，answer，desc，col，cat）; 
 mp.add（dp）; 
 placeMap.put（answer，dp）; 
  

NamedPlace和DescPlace都是Place的子类，我希望它们都在相同的HashMap中。

OP

我正在研究一个小项目。问题是我需要在项目的这个部分使用HashMap而不是ArrayList，因为HashMap对于搜索来说快很多。我创建了一个像这样的HashMap：

  HashMap< String，Object> placeMap = new HashMap<>（）; 
  

字符串是对象的名称，但事情是多个对象可以拥有一样的名字。所以我在我的搜索字段中搜索一个对象，并且我想将所有那些具有该名称的对象存储到一个ArrayList中，这样我就可以在它们中更改信息。

对象有很多不同的值，如名称，位置，一些布尔值等。

我需要在我的对象类中创建一个HashCode方法，它应该创建一个唯一的哈希码？

使用标准 Map< String，List< YourClassHere>>

解决方案 >实例，重要的是要记住，每个条目的映射值都是 List< YourClassHere> ，并且不会以任何特殊方式处理它。所以在你的情况下，如果你有

 私人地图< String，List< Place>> placeMap = new HashMap<>（）; 
  

然后存储您需要执行的值，如下所示：

  NamedPlace p = new NamedPlace（x，y，answer，col，cat）; 
列表< Place> list = placeMap.get（answer）; 
 list.add（p）; 
  

然而，这段代码有一些基本问题。



• 没有考虑到 answer 可能不存在于 placeMap 。


• 假设总是有一个 List< Place> 每个键你查询。
  
  
  

  所以解决这些潜在问题的最佳方法是做如下（Java 7和更高版本）：

/ p>



  NamedPlace p = new NamedPlace（x，y，answer，col，cat）; 
 
 if（placeMap.containsKey（answer）&& placeMap.get（answer）！= null）{
 placeMap.get（answer）.add（p）; 
} else {
列表< Place> list = new ArrayList< Place> （）; // ..或者无论你需要什么List实现
 list.add（p）; 
 placeMap.put（answer，list）; 
} 
  

如果您想扫描地点列表，代码会看起来像这个：



  if（placeMap.containsKey（key）&& placeMap.get（answer）！= null）{$ （Place p：placeMap.get（key））{
 //做东西
} 
} 
   p> 
EDIT

I've tried this HashMap with multiple values under the same key, and my hashMap now looks like this HashMap<String, List<Place>> placeMap = new HashMap<>();

Also tried to put Object instead of Place(place is my superclass). But when I now create my subclasses and wants to add them to the HashMap I get: 

  
The method put(String, List) in the type HashMap<String,List<Place>> is not applicable for the arguments (String, NamedPlace)
and 

  
The method put(String, List) in the type HashMap<String,List<Place>> is not applicable for the arguments (String, DescPlace)
here is my adding which created the error: 
NamedPlace p = new NamedPlace(x,y,answer,col,cat);
                placeMap.put(answer, p);




DescPlace dp = new DescPlace(x,y,answer, desc, col, cat);
                mp.add(dp);
                placeMap.put(answer, dp);
NamedPlace and DescPlace are both subclasses to Place, and I want them both in the same HashMap..

OP

I'm working on a little project here. The thing is that I need to use a HashMap instead of a ArrayList on this part of the project because HashMap is alot faster for searching. I've created a HashMap like this:
HashMap<String, Object> placeMap = new HashMap<>(); 
The String is the name of the Object, but the thing is that more than one object can have the same name. So I search for a object in my searchfield and I want to store all those objects that has that name into an ArrayList so I can change info in just them.

The object have alot of different values, like name, position, some booleans etc.

Do I need to create a HashCode method into my object class which shall create a unique hashcode?
解决方案
When using a standard Map<String, List<YourClassHere>> instance, it is important to remember that the map's values for each entry will be a List<YourClassHere>, and will not handle it in any special way. So in your case, if you have
private Map<String, List<Place>> placeMap = new HashMap<>();
Then to store values you will need to do as follows:
NamedPlace p = new NamedPlace(x,y,answer,col,cat);
List<Place> list = placeMap.get (answer);
list.add(p);
However, this piece of code has some underlying problems. 


It doesn't take into account that answer might not be present in placeMap.
It assumes that there's always a List<Place> instance for each key you query.


So the best way to fix those potential problems is to do as follows (Java 7 and later):
NamedPlace p = new NamedPlace(x,y,answer,col,cat);

if (placeMap.containsKey (answer) && placeMap.get (answer) != null) {
    placeMap.get (answer).add(p);
} else {
    List<Place> list = new ArrayList<Place> (); // ..or whatever List implementation you need
    list.add (p);
    placeMap.put (answer, list);
}
If you want to scna through the list of places, the code would look like this:
if (placeMap.containsKey (key) && placeMap.get (answer) != null) {
    for (Place p: placeMap.get (key)) {
        // Do stuff
    }
}


                        
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