以下是我的学生班

class Student implements Comparable {
   String name;
   int rollNo;

   @Override
   public int compareTo(Object obj) {
        return ((Student)obj).name.compareTo(this.name);
   }
} 

最新修改：但仍然没有得到正确的结果

@Override
public int compareTo(Object obj) {
    Student s = (Student) obj;
    if (name.equals(s.name)) { // achieving uniqueness
        return 0;
    } else {
        if (rollNo < s.rollNo) {
            return -1;
        } else if (rollNo > s.rollNo) {
            return 1;
        } else {
            // this makes `name` the second ordering option.
            // names don't equal here
            return name.compareTo(s.name);
        }
    }
}

如果我创建了TreeSet< Student>的对象,我将根据唯一名称&获取Student对象的排序列表.按名称排序.

但我需要在TreeSet中使用唯一的学生姓名< Student>按学生顺序排序.

比较器可以吗？任何人都可以帮助我,每个建议都表示赞赏.
谢谢.

更新：这是完整的程序：

public class Student implements Comparable {

    int rollNo;
    String name;

    Student(String n,int rno) {
        rollNo=rno;
        name=n;
    }

    /**
     * @param args
     */
    public static void main(String[] args) {

        TreeSet<Student> ts = new TreeSet<Student>();
        ts.add(new Student("bbb",2));
        ts.add(new Student("aaa",4));
        ts.add(new Student("bbb",2));
        ts.add(new Student("ccc",3));
        ts.add(new Student("aaa",1));
        ts.add(new Student("bbb",2));
        ts.add(new Student("bbb",5));

        System.out.println(ts);

    }

    @Override
    public int compareTo(Object obj) {
        Student s = (Student) obj;
        if (name.equals(s.name)) { // achieving uniqueness
            return 0;
        } else {
            if (rollNo < s.rollNo) {
                return -1;
            } else if (rollNo > s.rollNo) {
                return 1;
            } else {
                // this makes `name` the second ordering option.
                // names don't equal here
                return name.compareTo(s.name);
            }
        }
    }

    @Override
    public String toString() {
        return name + rollNo;
    }
}

更新：2：谢谢大家的建议,我还需要更多:)

/*
 * Actual scenario is having different properties,
 * So here I am just relating my actual scenario with Student class
 */
class Student implements Comparable {
    // sorting required on rollNo
    int rollNo;
    // Unique name is required
    String name;

    Student(String n, int rno) {
        rollNo = rno;
        name = n;
    }

    /**
     * 
     * @param args
     */
    public static void main(String[] args) {

        TreeSet<Student> tsName = new TreeSet<Student>();
        // here by default, order & uniqueness by name only
        tsName.add(new Student("ccc", 2));
        tsName.add(new Student("aaa", 4));
        tsName.add(new Student("ddd", 1));
        tsName.add(new Student("bbb", 3));
        tsName.add(new Student("ddd", 5));
        // output: aaa:4, bbb:3, ccc:2, ddd:1
        System.out.println(tsName);

        // creating new comparator for student RollNo
        TreeSet<Student> tsRollNo = new TreeSet<Student>(new Comparator<Student>() {
                    public int compare(Student stud1, Student stud2) {
                        return new Integer(stud1.rollNo).compareTo(stud2.rollNo);
                    }
                });
        tsRollNo.addAll(tsName);
        System.out.println(tsRollNo);
        // now got the desire output: ddd:1, ccc:2, bbb:3, aaa:4
    }

    public boolean equals(Object obj) {
        // internally not used to check equality while adding objects
        // in TreeSet
        System.out.println("equals() for " + this + " & " + ((Student) obj));
        return false;// return false/true doesn't make any sense here
    }

    @Override
    public int compareTo(Object obj) {
        Student s = (Student) obj;
        // internally inside TreeSet, compareTo is used to decide
        // whether two objects are equal or not,
        // i.e. compareTo will return 0 for same object(here student name)
        System.out.println("compareTo() for " + this + " & " + ((Student) obj));
        // achieving uniqueness
        return name.compareTo(s.name);
    }

    @Override
    public String toString() {
        return name + ":" + rollNo;
    }
}

OUTPUT：

compareTo() for aaa:4 & ccc:2
compareTo() for ddd:1 & ccc:2
compareTo() for bbb:3 & ccc:2
compareTo() for bbb:3 & aaa:4
compareTo() for ddd:5 & ccc:2
compareTo() for ddd:5 & ddd:1
[aaa:4, bbb:3, ccc:2, ddd:1]
[ddd:1, ccc:2, bbb:3, aaa:4]

朋友,无论我使用两个比较器得到什么,是否有可能
添加对象时实现相同的??
我不能先添加元素&然后使用新的比较器来实现所需的顺序.
我正在操纵数以千计的价值,因此也需要考虑性能.

解决方法:

在TreeSet中它将使用比较器,同时添加用于排序和唯一检查的元素,

现在的问题是,如果你使用比较器滚动否,你将按照滚动号和唯一滚动nos进行排序.你不能把两者放在一起.

我建议你去.

> TreeSet在这里集中关于重复删除
>然后,一旦获得了唯一数据,就可以使用ArrayList并按照您想要的任何顺序对其进行排序