使用 Java 的 DSA - 排序对象
可以使用 java.util.Arrays.sort() 方法轻松对 Java 对象进行排序。请考虑以下示例。
ObjectSortingDemo.java
package com.tutorialspoint.simplesort;
public class ObjectSortingDemo {
public static void main(String[] args){
/* String 对象数组 */
String[] employees
= new String[] {"Robert", "Paul","John","Micheal" };
System.out.println("未排序数组 : "
+ Arrays.toString(employees));
/* 按字典顺序对数组进行排序 */
Arrays.sort(employees);
System.out.println("已排序数组:"
+ Arrays.toString(employees));
System.out.println();
}
}
如果我们编译并运行上述程序,则会产生以下结果:
未排序数组: [Robert, Paul, John, Micheal] 已排序数组 : [John, Micheal, Paul, Robert]
使用 Comparable 接口
为了对对象进行排序,其类可以实现 java.lang.Comparable 接口。请考虑以下代码。
Employee.java
package com.tutorialspoint.simplesort;
public class Employee implements Comparable<Employee> {
private int employeeId;
private String name;
private String department;
public Employee (int employeeId,String name, String department){
this.employeeId = employeeId;
this.name = name;
this.department = department;
}
@Override
public int compareTo(Employee employee) {
return employeeId - employee.employeeId;
}
public String getName(){
return name;
}
public String getDepartment(){
return department;
}
@Override
public String toString() {
return "
[ " + employeeId
+"," + name
+"," + department
+" ]";
}
}
此处 Employee 类实现 java.lang.Comparable 接口并具有方法 compareTo()。Array.sort() 使用合并排序算法并使用此 compareTo() 方法比较两个对象,以便对作为参数传递的对象数组进行排序。
ObjectSortingDemo.java
public class ObjectSortingDemo {
public static void main(String[] args){
/* 使用 Arrays.sort() 方法对数组进行排序 */
Employee[] employeesObjects = new Employee[] {
new Employee(100, "Robert", "Finance"),
new Employee(30, "Paul", "Finance"),
new Employee(50, "John", "Finance"),
new Employee(12, "Micheal", "Finance")
};
System.out.println("Unsorted Array : "
+ Arrays.toString(employeesObjects));
Arrays.sort(employeesObjects);
System.out.println();
System.out.println("Sorted Array by id: "
+ Arrays.toString(employeesObjects));
}
}
如果我们编译并运行上述程序,则会产生以下结果 −
Unsorted Array : [ [ 100,Robert,Finance ], [ 30,Paul,Finance ], [ 50,John,Finance ], [ 12,Micheal,Finance ]] Sorted Array by id: [ [ 12,Micheal,Finance ], [ 30,Paul,Finance ], [ 50,John,Finance ], [ 100,Robert,Finance ]]
使用 Comparator 接口
使用 java.util.Comparator 接口让我们能够精确控制对象的排序。正如我们在前面的例子中看到的,我们通过实现 java.lang.Comparable 接口设置了一个标准,即员工应该根据类中的 compareTo() 方法进行排序。使用 Comparator 类,如果类没有实现可比较的接口,我们可以在不修改类的情况下设置标准。请考虑以下代码。
Employee.java
package com.tutorialspoint.simplesort;
public class Employee {
private int employeeId;
private String name;
private String department;
public Employee (int employeeId,String name, String department){
this.employeeId = employeeId;
this.name = name;
this.department = department;
}
public String getName(){
return name;
}
public String getDepartment(){
return department;
}
@Override
public String toString() {
return "
[ " + employeeId
+"," + name
+"," + department
+" ]";
}
}
定义一个比较器,可以根据姓名比较两个员工。
EmployeeNameComparator.java
package com.tutorialspoint.simplesort;
import java.util.Comparator;
public class EmployeeNameComparator implements Comparator<Employee> {
@Override
public int compare(Employee employee1, Employee employee2) {
return employee1.getName().compareTo(employee2.getName());
}
}
此处 EmployeeNameComparator 类实现 java.util.Comparator 接口并具有方法 compare()。Array.sort() 使用合并排序算法并使用此 compare() 方法比较两个对象,以便对作为参数传递的对象数组进行排序。
ObjectSortingDemo.java
public class ObjectSortingDemo {
public static void main(String[] args){
/* 使用 Arrays.sort() 方法对数组进行排序 */
Employee[] employeesObjects = new Employee[] {
new Employee(100, "Robert", "Finance"),
new Employee(30, "Paul", "Finance"),
new Employee(50, "John", "Finance"),
new Employee(12, "Micheal", "Finance")
};
System.out.println("Unsorted Array : "
+ Arrays.toString(employeesObjects));
Arrays.sort(employeesObjects,new EmployeeNameComparator());
System.out.println();
System.out.println("Sorted Array by name: "
+ Arrays.toString(employeesObjects));
}
}
如果我们编译并运行上述程序,则会产生以下结果 −
Unsorted Array : [ [ 100,Robert,Finance ], [ 30,Paul,Finance ], [ 50,John,Finance ], [ 12,Micheal,Finance ]] Sorted Array by name: [ [ 50,John,Finance ], [ 12,Micheal,Finance ], [ 30,Paul,Finance ], [ 100,Robert,Finance ]]

