需要编写Manager和Employee类的代码。如何在不破坏封装的情况下使它们彼此“可见”？

Question

Manager和Employee类都是EnterpriseMember的子类。如何为Employee类编写“getManager”方法（返回在其报表列表中包含此Employee的Manager实例）？提前致谢！

public class Manager extends EnterpriseMember {

    /*Fields */
    private List reports = new ArrayList();

    /*Constructor */
    public Manager(String name){
        super(name);
    }

    /*Methods */
    public void addReport(Employee employee){            
        reports.add(employee);
    }// How can "employee" know it is in this List?

}

public class Employee extends EnterpriseMember {

    /*Constructor */
    public Manager(String name){
        super(name);
    }

    /*Methods */
    public Manager getManager(){            
        return ???;
    }

}

Answer 1

像这样的东西：

public class Manager {

    private List<Employee> reports = new ArrayList<Employee>();

    public void addReport(Employee e) {
        if (e != null) {
            this.reports.add(e);
            e.setManager(this);
        }
    }
}

public class Employee {

    private Manager manager;

    public void setManager(Manager m) {
        if (m != null) {
            this.manager = m;
        }
    }
}

如果不清楚，您应该添加所需的所有其他方法。我只说明了如何更新Manager中的Employee引用，当它被添加到直接报告的List时。

你还应该有一个removeReport方法，从Employee中删除List并将其Manager设置为null。

你打算如何在这个Employee找到一个List？按名字？员工ID？提示：考虑适当地为你的课程重写equals和hashCode。

是不是Managers也Employees？老板有没有老板？这是一个层次结构，一棵树。

Answer 2

通常，具有不同属性的对象如下所示：

public class Employee extends EnterpriseMember {

    private Manager manager;
    private String name; // You probably don't need this because you defined it in the Superclass.
    .
    .
    .

    /*Constructor */
    public Employee(String name){
        super(name);
    }

    /*Methods */
    public Manager getManager(){            
        return manager;
    }

    public void setManager(Manager manager){            
        this.manager = manager
    }

    // Other getters and setters for the attributes.

}