Java 将 ArrayList 转换为有序集合（TreeSet）并返回
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Converts ArrayList into a sorted set (TreeSet) and returns it
提问by Good tree
I a method that takes a list of numbers (e.g. ArrayList) and converts it into a sorted set (e.g. TreeSet) and returns it. I wrote code, but I'm having some problems.
我是一个方法，它接受一个数字列表（例如ArrayList）并将其转换为一个有序集合（例如TreeSet）并返回它。我写了代码，但我遇到了一些问题。
public TreeSet getSort (ArrayList list){
TreeSet set =new TreeSet(list);
return set;
My problem is in main:
我的问题主要是：
ArrayList List = new ArrayList();
List.add(5);
List.add(55);
List.add(88);
List.add(555);
List.add(154);
System.out.println("the TreeSet of ArrayList is : " + getSort(List));
采纳答案by Nambi
You need to have the class instance to call the getSort()method or make the getSort()to static
你需要有类的实例来调用getSort()方法，或使getSort()以static
Do like
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System.out.println("the TreeSet of ArrayList is : "+ new classname().getSort(List));
or make the method static
或使方法静态
public static TreeSet getSort (ArrayList list){
TreeSet set =new TreeSet(list);
return set;
}
回答by arshajii
You're probably getting an error because getSort()isn't static, so it can't be called from main(). Beyond this, you shouldn't use raw types, parametrize your lists and sets:
您可能会收到错误，因为getSort()is not static，因此无法从 调用它main()。除此之外，你不应该使用原始类型，参数化你的列表和集合：
ArrayList<Integer> list = ...
TreeSet<Integer> set = ...
You should be getting warnings about this.
您应该收到有关此的警告。
In fact, I would make this method totally generic:
事实上，我会让这个方法完全通用：
public static <V extends Comparable<V>> TreeSet<V> getSort(List<V> list) {
return new TreeSet<>(list);
}
Lastly, remember to follow naming conventions: local variable names should start with a lowercase letter (i.e. listand not List).
最后，请记住遵循命名约定：局部变量名称应以小写字母开头（即list而不是List）。
回答by Anubian Noob
getSort()isn't static and can't be called from main. You need to make it static.
getSort()不是静态的，不能从主调用。你需要让它静态。
