关于set

public static void main(String[] args) {
        //创建一个集合set对象
        //Set<Student> set = new TreeSet<Student>();
        //Set<Student> set = new HashSet<Student>();
        Set<Student> set = new LinkedHashSet<Student>();
        //添加多个学生
        Student stu2 = new Student(2, "lisi", 23, 98);
        Student stu3 = new Student(3, "wangwu", 22, 87);
        Student stu1 = new Student(1, "zhangsan", 23, 90);
        Student stu4 = new Student(1, "zhangsan", 23, 90);
        set.add(stu1);
        set.add(stu2);
        set.add(stu3);
        set.add(stu4);
        //输出学生
        System.out.println(set.size());
        System.out.println(set);
    }

问题1：HashSet、LinkedHashSet :为什么String有重复，会保持唯一；为什么Student有重复，不会保持唯一。

问题2：TreeSet  为什么String可以添加，而Student就不让添加到TreeSet中呢？ 而是抛出异常：

 

java.lang.ClassCastException: com.bjsxt.entity.Student cannot be cast to java.lang.Comparable

思考：String是系统类，Student是自定义类，应该是String已经做了某些事情，但是Student没有做

解答1：HashSet、LinkedHashSet 需要Student实现hashCode()和equals()

解答2：TreeSet 需要Student实现Comparable接口并指定比较的规则

 

public class Student implements Comparable<Student>{
        private int sno;
        private String name;
        private int age;
        private double score;
        @Override
        public int compareTo(Student o) {
            //return this.sno - o.sno;
            //return o.sno - this.sno;
            return -(this.sno - o.sno);
        }
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Student student = (Student) o;
            if (sno != student.sno) return false;
            if (age != student.age) return false;
            if (Double.compare(student.score, score) != 0) return false;
            return name != null ? name.equals(student.name) :
                    student.name == null;
        }
        @Override
        public int hashCode() {
            int result;
            long temp;
            result = sno;
            result = 31 * result + (name != null ? name.hashCode() : 0);
            result = 31 * result + age;
            temp = Double.doubleToLongBits(score);
            result = 31 * result + (int) (temp ^ (temp >>> 32));
            return result;
        }
    }