java - 编写一个从 arraylist 返回迭代器的方法

Question

已关闭。此问题不符合 Stack Overflow guidelines 。目前不接受答案。

---

要求提供代码的问题必须表现出对所解决问题的最低限度的了解。包括尝试的解决方案、为什么它们不起作用以及预期结果。另请参阅:Stack Overflow question checklist

已关闭 9 年前。

Improve this question

我正在使用 blueJ 来编写此内容。我想做的是 ddd 一个名为 getUserIterator() 的方法至UserGroup那returns an Iterator来自ArrayList并使用 Iterator我的方法返回以循环现有用户并添加 userType 的任何用户是 "admin"到管理员组。我不知道该怎么做!我的代码如下:

用户类别:

package user; 
public class User{
    public enum UserType{                          
        ADMIN, EDITOR, USER;
    }

    private String id;                            
    private UserType userPermissions;              
    private String actualName;                     

    public User(String username, UserType userType, String name){
        id = username;
        userPermissions = userType;
        actualName= name;
    }

    public String getUsername(){
        return id;
    }

    public UserType getUserType(){
        return userPermissions;
    }       

    public String getName(){
        return actualName;
    }

    public void setUserType(UserType input){
        userPermissions = input;
    }
}

userGroup 类是:

package user;
import java.util.*;
import user.User.UserType; 

public class UserGroup{

    private ArrayList<User> people;

    public UserGroup(){
        people = new ArrayList<User>();
    }

    public void addSampleData(){
        people.add(new User("jar1g13", UserType.ADMIN,"Jonny"));
        people.add(new User("ao9", UserType.EDITOR,"Aniruddh"));
        people.add(new User("pe6", UserType.USER,"Peter"));
        people.add(new User("mat73", UserType.USER,"Matthew"));
        people.add(new User("ora69", UserType.EDITOR,"Oranthi"));
        people.add(new User("ben12", UserType.USER,"Benedict"));
        people.add(new User("cam30", UserType.ADMIN,"Cambyse"));
        people.add(new User("are20", UserType.USER,"Alex"));
        people.add(new User("lim19", UserType.USER,"Liam"));
        people.add(new User("ada13", UserType.EDITOR,"Adam"));
    } 

    public User getUser(int idx){
        return people.get(idx);
    }

    public void printUsernames(){
        for (User user: people){
            System.out.printf("%s %s\n", user.getUsername(), user.getUserType());
        }
    }

    public void removeFirstUser(){
        people.remove(0);
    }

    public void removeLastUser(){
        people.remove(people.size()-1);
    }

    public void removeUser(String username){
        Iterator<User> itr = people.iterator();
        while (itr.hasNext()){
            User element = (User) itr.next();
            if (element.getUsername().equals(username)){
                itr.remove();                                       
            }
        }
    }

}

主类:

package user;

public class Main{

    public static void main(String[] args){
        UserGroup grp = new UserGroup();          
        grp.addSampleData();                       
        grp.printUsernames();                      
    }

    public static void admins(){
        UserGroup administrators - new UserGroup(); 
    }
}

Answer 1

我会把它写成

public class UserGroup implements Iterable<User> {
    private final List<User> people = new ArrayList<User>();

    public Iterator<User> iterator() {
        return people.iterator();
    }
    // OR to prevent the list being modified.
    public Iterator<User> iterator() {
        return Collections.unmodifableList(people).iterator();
    }
    // OR for extra paranoia
    public Iterator<User> iterator() {
        return Collections.unmodifableList(new ArrayList<Person>(people)).iterator();
    }
    // OR for even more paranoia
    public Iterator<User> iterator() {
        List<People> ret = new ArrayList<People>();
        for(People p : people)
            ret.add(new Person(p)); // take a deep copy.
        return Collections.unmodifableList(ret).iterator();
    }
}

// later
UserGroup userGroup = ...
for(User u : userGroup) {
   // do something with u
}

这假设您不需要担心调用者修改列表(这是永远不应该做的，但有些人担心这样的事情)