用Collections.sort方法对list排序有两种方法

第一种是list中的对象实现Comparable接口，如下：
/**
* 根据order对User排序
*/
public class User implements Comparable<User>{
    private String name;
    private Integer order;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Integer getOrder() {
        return order;
    }
    public void setOrder(Integer order) {
        this.order = order;
    }
    public int compareTo(User arg0) {
        return this.getOrder().compareTo(arg0.getOrder());
    }
}
测试一下：
public class Test{

    public static void main(String[] args) {
        User user1 = new User();
        user1.setName("a");
        user1.setOrder(1);
        User user2 = new User();
        user2.setName("b");
        user2.setOrder(2);
        List<User> list = new ArrayList<User>();
        //此处add user2再add user1
        list.add(user2);
        list.add(user1);
        Collections.sort(list);
        for(User u : list){
            System.out.println(u.getName());
        }
    }
}
输出结果如下
a
b
第二种方法是根据Collections.sort重载方法来实现，例如：
/**
* 根据order对User排序
*/
public class User { //此处无需实现Comparable接口
    private String name;
    private Integer order;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Integer getOrder() {
        return order;
    }
    public void setOrder(Integer order) {
        this.order = order;
    }
}
主类中这样写即可：
public class Test{
    public static void main(String[] args) {
        User user1 = new User();
        user1.setName("a");
        user1.setOrder(1);
        User user2 = new User();
        user2.setName("b");
        user2.setOrder(2);
        List<User> list = new ArrayList<User>();
        list.add(user2);
        list.add(user1);
       
        Collections.sort(list,new Comparator<User>(){
            public int compare(User arg0, User arg1) {
                return arg0.getOrder().compareTo(arg1.getOrder());
            }
        });
        for(User u : list){
            System.out.println(u.getName());
        }
    }
}
输出结果如下
a
b
前者代码结构简单，但是只能根据固定的属性排序，后者灵活，可以临时指定排序项，但是代码不够简洁

多字段的场合：
Collections.sort(list,new Comparator<User>(){
            public int compare(User arg0, User arg1) {
//            第一次比较专业
                int i = arg0.getOrder().compareTo(arg1.getOrder());

//            如果专业相同则进行第二次比较
            if(i==0){
//                第二次比较
                int j=arg0.getXXX().compareTo(arg1.getXXX());
//                如果学制相同则返回按年龄排序
                if(j==0){
                    return arg0.getCCC().compareTo(arg1.getCCC());
                }
                return j;
            }
            return i;
            }
        });


以下是另外一个例子
package test;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

public class ArrayListTest {
public static void main(String[] args) {
ArrayList<Exmployee> arrayList  = new ArrayList<Exmployee>() ;
arrayList.add(new Exmployee("zengqiang",new Integer(5000))) ;
arrayList.add(new Exmployee("zengmin",new Integer(4000))) ;
arrayList.add(new Exmployee("liuxiaojuan",new Integer(4200))) ;
arrayList.add(new Exmployee("giuming",new Integer(2200))) ;
Collections.sort(arrayList, new Comparator<Exmployee>(){
public int compare(Exmployee arg0, Exmployee arg1) {
return  arg1.getSalary()-arg0.getSalary() ; //按照工资升序
//return  arg0.getSalary()-arg1.getSalary() ; 按照工资降序
//return arg0.getName().compareTo(arg1.getName()) ; 按照姓名升序
            }
});
for(Exmployee e:arrayList)
System.out.println(e.getName()+"'s salary is "+e.getSalary()) ;
}

}

class Exmployee {
public Exmployee(String name, int salary) {
this.name = name ;
this.salary = salary ;
}

private String name;

public String getName() {
return name;
}

public void setName(String name) {
this.name = name;
}

public int getSalary() {
return salary;
}

public void setSalary(int salary) {
this.salary = salary;
}

private int salary;
}

评论

3 楼 xiangxixids01 2010-07-26  

这个问题, 我也纠结了很久, 包括list的本地分页等问题. 不过最终还是解决了, 排序的部分, 感谢楼主的帖子~
PS: 当数据量比较大时, java处理速度如何呢?
比如接近5W条记录的时候.

2 楼 justdo2008 2010-06-25  

第二种方式貌似实现Serializable接口才能使用。

1 楼 tmartin 2010-06-19  

感谢啊 解决我问题了