有的时候,我们需要对获取的list集合进行排序,然后输出。那么我们一般会用到Collections.sort。
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:
[java] view plain copy
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/**
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* 根据order对User排序
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*/
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public class User implements Comparable<User>{
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private String name;
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private Integer order;
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public String getName() {
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return name;
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}
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public void setName(String name) {
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this.name = name;
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}
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public Integer getOrder() {
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return order;
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}
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public void setOrder(Integer order) {
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this.order = order;
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}
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public int compareTo(User arg0) {
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return this.getOrder().compareTo(arg0.getOrder());
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}
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} 测试一下:
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public class Test{
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public static void main(String[] args) {
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User user1 = new User();
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user1.setName("a");
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user1.setOrder(1);
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User user2 = new User();
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user2.setName("b");
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user2.setOrder(2);
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List<User> list = new ArrayList<User>();
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//此处add user2再add user1
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list.add(user2);
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list.add(user1);
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Collections.sort(list);
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for(User u : list){
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System.out.println(u.getName());
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}
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}
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}
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:
[java] view plain copy
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/**
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* 根据order对User排序
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*/
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public class User { //此处无需实现Comparable接口
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private String name;
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private Integer order;
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public String getName() {
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return name;
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}
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public void setName(String name) {
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this.name = name;
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}
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public Integer getOrder() {
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return order;
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}
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public void setOrder(Integer order) {
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this.order = order;
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}
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}
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主类中这样写即可(HastSet——>List——>sort进行排序):
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public class Test {
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public static void main(String[] args) {
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User user1 = new User();
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user1.setName("a");
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user1.setOrder(1);
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User user2 = new User();
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user2.setName("b");
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user2.setOrder(2);
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Set<User> Hset = new HashSet<User>(); //set不能存放重复的对象
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Hset.add(user2);
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Hset.add(user1);
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List<User> list = new ArrayList<User>();
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list.addAll(Hset);
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Collections.sort(list,new Comparator<User>(){
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public int compare(User arg0, User arg1) {
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return arg1.getOrder().compareTo(arg0.getOrder());
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}
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});
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for(User u : list){
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System.out.println(u.getName());
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}
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}
输出结果如下:
a
b
默认为升序,将。return arg0.getOrder().compareTo(arg1.getOrder()); 改为:
return arg1.getOrder().compareTo(arg0.getOrder());
就成降序的了。
本文转自建波李 51CTO博客,原文链接:http://blog.51cto.com/jianboli/1890628,如需转载请自行联系原作者