在Java中,如何快速排序对象的ArrayList,其中排序字段是多层深度？

Question

基本上,我有一个名为"Employees"的Container类,其中包含一个ArrayList.此ArrayList包含"Employee"对象,后者又包含"EmployeeData"对象,而这些对象又包含String对象,例如"first"或"last"(这是员工姓名).

这是ArrayList结构的图表:

ArrayList[Employee] emps ==> 1:Many ==> Employee emp
Employee emp ==> 1:1 ==> EmployeeData data
EmployeeData data ==> 1:2 ==> String last // A string that contains employee's last name.

我将如何在ArrayList上执行快速排序,以便其中的"Employee"对象基于String对象"last"按字母顺序排列？看起来有点复杂!

---

这是我的课程的基本设计:

class Employees{
    //data:
        private ArrayList<Employee> emps = new ArrayList<Employee>();

    //Some constructors go here

    //Methods to add, remove, toString, etc, go here

    public /*output a sorted ArrayList?*/ sort(){
        // Some kind of "quicksort" in here to modify or create a new ArrayList sorted by employee's las name...
    }
}

class Employee{
    //data:
    EmployeeData data;
    // Some methods to construct and modify EmployeeData data.
}

class EmployeeData{
    //data:
        String first, last; // I wish to sort with "last". How do you do it?
        double payrate, hours;
    //...methods...
}

如您所见,这些是课程.我不知道如何在"Employees"类中实现"sort",以便它通过"EmployeeData"类的"last"变量对ArrayList进行排序.

Answer 1

你可以做一个比较器,如:

public class MyComparator implements Comparator<Employee>
{
  public int compare(Employee e1, Employee e2)
  {
    return e1.getData().getLast().compareTo(e2.getData().getLast());
  }
}

然后用它来排序列表.

Collections.sort(myList, new MyComparator());

或者,您可以使用TreeSet对使用此比较器的插入进行排序,或者使Employee成为可比较的对象,以使用Collections或SortedSet进行排序.

public class Employee implements Comperable<Employee>
{
  ...
  public int compareTo(Employee e)
  {
    return this.getData().getLast().compareTo(e.getData().getLast());
  }
  ...
}