通过一个实例讲解 Collections.sort() 用法 发表于 2019-07-23 | 阅读次数: Collections.sort() 方法应该是开发中最常用的排序方法吧! 下面通过一个例子讲解简单用法: 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667// 首先定义一个 POJOpublic class User { String name; String age; public User(String name, String age) { this.name = name; this.age = age; } public String getAge() { return age; } public void setAge(String age) { this.age = age; } public String getName() { return name; } public void setName(String name) { this.name = name; }}// 再定义 POJO 的比较类 import java.util.Comparator;public class UserComparator implements Comparator { public int compare(Object arg0, Object arg1) { User user0 = (User) arg0; User user1 = (User) arg1; // 首先比较年龄,如果年龄相同,则比较名字 int flag = user0.getAge().compareTo(user1.getAge()); if (flag == 0) { return user0.getName().compareTo(user1.getName()); } else { return flag; } }}// 测试方法 public static void main(String[] args) { List userlist = new ArrayList(); userlist.add(new User("dd", "4")); userlist.add(new User("aa", "1")); userlist.add(new User("ee", "5")); userlist.add(new User("bb", "2")); userlist.add(new User("ff", "5")); userlist.add(new User("cc", "3")); userlist.add(new User("gg", "6")); UserComparator comparator = new UserComparator(); Collections.sort(userlist, comparator); for (int i = 0; i < userlist.size(); i++) { User user_temp = (User) userlist.get(i); System.out.println(user_temp.getAge() + "," + user_temp.getName()); } } 结果: 相关文章 从字节码角度探究 JDK 和 CGLIB 动态代理区别 Guava 有限大小队列 (EvictingQueue) 介绍和分析 闲谈 Tomcat 性能优化,通过 ExpiresFilter 设置浏览器缓存 使用 Collections.unmodifiableMap() 方法,实现返回不可修改的集合。 Tomcat:Invalid character found in the request target. The valid characters are defined in RFC 7230 and RFC 3986 本文链接: https://zhangzw.com/posts/20190723.html 版权声明: 本博客所有文章除特别声明外,均采用 BY-NC-ND 许可协议。转载请注明出处!