Comparator 排序时去掉了重复元素

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]]