返回ArrayList字符串的方法

So I have:

所以我有:

private static ArrayList<AbstractAnalyser> analysers = new ArrayList<>();

public static String getAnalyser(String analyser){
    if(analysers.contains(analyser)){
      return "The full name of the analyser";
    }
    return null;
}

So what I want is..

所以我想要的是......

If the the arraylist contains the parameter analyser, I want it to return the full name of the object what is in the arraylist.

如果arraylist包含参数分析器,我希望它返回arraylist中对象的全名。

Let's say these values are in the arraylist:

假设这些值在arraylist中:

analyser, method and second. <- random names

分析仪,方法和第二。 < - 随机名称

If the parameter is analyser and the arraylist contains analyser. The method needs the return that name.

如果参数是分析器,则arraylist包含分析器。该方法需要返回该名称。

Even when the parameter is "analy".

即使参数是“分析”。

1 个解决方案

#1

I would use a Map instead of a List:

我会使用Map而不是List:

private static Map<String, AbstractAnalyser> analysers = new HashMap<>();

public static AbstractAnalyser getAnalyser(String analyserName){
    AbstractAnalyser result = null;
    if ((analyserName != null) && (analyserName.trim().length() > 0)) {
        if (analysers.containsKey(analyserName)) {
            result = analysers.get(analyserName);
        } else {
            for (String key : analysers.keySet()) {
                // put the logic to find the one you want here.       
            }
        }
    }
    return result;
}

But, if you must, you can do it this way if the AbstractAnalyser has a way to give you its name:

但是,如果必须,如果AbstractAnalyser能够为您提供名称,则可以这样做:

private static List<AbstractAnalyser> analysers = new ArrayList<>();

public static AbstractAnalyser getAnalyser(String analyserName){
    AbstractAnalyser result = null;
    if ((analyserName != null) && (analyserName.trim().length() > 0)) {
        for (AbstractAnalyser analyser : analysers) {
            // Here's how you look by name
            if (analyser.getName().equals(analyserName)) {  
                result = analyser;
                break;    
            } else {
               // put special logic to find the one you want here. 
            }
        }
    }
    return result;
}

Using the Map is always more efficient when you give the exact name because the lookup is O(1). The List lookup is always O(N).

当您给出确切的名称时,使用Map总是更有效,因为查找是O(1)。列表查找始终为O(N)。

#1