public class User { private String name; private int age; public String getName() { return name; } public void setName(String name) { this.name = name; } public int getAge() { return age; } public void setAge(int age) { this.age = age; } @Override public String toString() { return "User [age=" + age + "]"; } }
public static void main(String[] args) { TreeSet<User> set = new TreeSet(new Comparator() { @Override public int compare(Object o1, Object o2) { return ((User) o2).getAge() - (((User) o1).getAge()); } }); User user1 = new User(); user1.setAge(10); User user2 = new User(); user2.setAge(20); User user3 = new User(); user3.setAge(10); set.add(user1); set.add(user2); set.add(user3); System.out.println(set.size()); }
你认为最终输出的size是多少?3还是2呢?
其实是2,因为两个age相等,compare(Object o1, Object o2) 方法返回值为0,所以认为是重复元素,没有将user3加入到集合.
我一直以为去重复是通过hashcode 和 equals方法来判断的,现在看来不是的.
这个方法排序是按compare(Object o1, Object o2)返回值从小到大排序, 如果返回值是0则认为两个对象是重复的(而不是并列排序),如果想要让其认为两个对象是并列的, 那么不应该返回0,而是返回非0的数字.
解决办法
1.在计算差值之前判断两个age是否相等,相等则返回非0的数字
public int compare(Object o1, Object o2) { if (((User) o2).getAge() == (((User) o1).getAge())) { return 1; } return ((User) o2).getAge() - (((User) o1).getAge()); }
上述修改会影响最终排序吗?不会!事例如下:
public static void main(String[] args) { TreeSet<User> set = new TreeSet(new Comparator() { @Override public int compare(Object o1, Object o2) { if (((User) o2).getAge() == (((User) o1).getAge())) { return 1; } return ((User) o2).getAge() - (((User) o1).getAge()); } }); User user1 = new User(); user1.setAge(10); User user2 = new User(); user2.setAge(20); User user3 = new User(); user3.setAge(10); User user4 = new User(); user4.setAge(15); User user5 = new User(); user5.setAge(15); User user6 = new User(); user6.setAge(12); User user7 = new User(); user7.setAge(12); set.add(user1); set.add(user2); set.add(user3); set.add(user4); set.add(user5); set.add(user6); set.add(user7); System.out.println(set); System.out.println(set.size()); }
打印结果如下:
[User [age=20], User [age=15], User [age=15], User [age=12], User [age=12], User [age=10], User [age=10]]
7
2.用Collections.sort排序
public static void main(String[] args) { List<User> list = new ArrayList<>(); User user1 = new User(); user1.setAge(10); User user2 = new User(); user2.setAge(20); User user3 = new User(); user3.setAge(10); list.add(user1); list.add(user2); list.add(user3); Collections.sort(list, new Comparator() { @Override public int compare(Object o1, Object o2) { return ((User) o2).getAge() - (((User) o1).getAge()); } }); System.out.println(list); }
打印如下:
[User [age=20], User [age=10], User [age=10]]