问题描述

重写 compareTo后，ArrayList.sort可以升序排列，为什么一般不改成降序呢 ？

package interfaces;
import java.util.*;
//实例数组排序：员工数组排序
/**

• This program demonstrates the use of the Comparable interface.
• @version 1.30 2004-02-27
• @author Cay Horstmann
  */
  public class EmployeeSortTest
  {
  public static void main(String[] args)
  {
  Employee[] staff = new Employee[3];

  staff[0] = new Employee("Harry Hacker", 35000);
  staff[1] = new Employee("Carl Cracker", 75000);
  staff[2] = new Employee("Tony Tester", 38000);

  Arrays.sort(staff);//结合重写后的compareTo方法，将类按照工资高低升序排列

  // print out information about all Employee objects
  for (Employee e : staff)
  System.out.println("name=" + e.getName() + ",salary=" + e.getSalary());
  }
  }

class Employee implements Comparable
{
private String name;
private double salary;

public Employee(String n, double s) {
    name = n;
    salary = s;
}

public String getName()
{
return name;
}

public double getSalary()
{
return salary;
}

public void raiseSalary(double byPercent)
{
double raise = salary * byPercent / 100;
salary += raise;
}

/**
* Compares employees by salary
* @param other another Employee object
* @return a negative value if this employee has a lower salary than
* otherObject, 0 if the salaries are the same, a positive value otherwise
*/
public int compareTo(Employee other)
{
if (salary < other.salary) return -1;//if (salary < other.salary) return -1;
if (salary > other.salary) return 1;//if (salary > other.salary) return 1;
return 0; //若上面两句改写成这样，那输出结果就会降序排列，也不会报错，为什么不这样写呢？
}

}

问题在注释中写了

解决方案

sort的源码中，判断方式是下面这种方式
if(下一个对象.compareTo(前一个对象)<0){
}else{
}
也就是说只判断了后一个对象比前一个对象小情况，当小的时候就是升序，其他都是else都是降序

所以你return 0和return 1是一个效果

解决方案二：

升序讲叙是完全对称的，从算法角度来说都一样，写哪个可以

解决方案三：

根据你的需要，升序用的多一些，仅此而已。

解决方案四：

都可以。。。无所谓的吧。。。。看需求