编辑：正如您所看到的,因为我正在使用的选择排序需要查找更大的值,所以我用大替换了lessOrEqual.效果是一样的,我这样做只是为了提高可读性.

然后,我实现了另一个静态函数,该函数接受ArrayList< Student&gt ;,对其进行排序并返回排序后的结果.所使用的排序算法非常基础：选择排序. O(N ^ 2)效率不高,但是为了简化起见,在下面的演示中我这样做了. 码：

import java.util.ArrayList; 

public class Student {
    private String name;
    private Integer rollNumber;
    private int m1;
    private int m2;
    private int m3;
    private int totMarks;

    public static boolean larger(Student s1,Student s2){
        if(s1.totMarks < s2.totMarks) return false; 
        else if (s1.totMarks > s2.totMarks) return true;
        // compare names
        else return s1.name.compareTo(s2.name) > 0;
    }

    public static ArrayList<Student> sortSelection(ArrayList<Student> list){
        for(int i=0; i<list.size(); i++){
            for(int j=i+1; j< list.size(); j++){
                if(larger(list.get(i),list.get(j))){ // swap
                    Student temp = list.get(i); 
                    list.set(i,list.get(j));
                    list.set(j,temp);
                }
            }
        }
        return list;
    }
    //Getter setter
    public String getName(){
        return name; 
    }

    public void setName(String name){
        this.name = name; 
    }

    public int getTotMarks(){
        return totMarks;
    }

    public void setTotMarks(int totMarks){
        this.totMarks = totMarks; 
    }

    @Override
    public String toString(){
        return String.format("Name: %20s - Total Marks: %3d",name,totMarks);
    }

    public static void main(String[] args){
        Student s1 = new Student(); 
        Student s2 = new Student();
        Student s3 = new Student();
        Student s4 = new Student();

        s1.setName("John Smith");
        s1.setTotMarks(98);
        s2.setName("Jack Smith");
        s2.setTotMarks(98);
        s3.setName("Adam Noob");
        s3.setTotMarks(100);
        s4.setName("Ved Parkash");
        s4.setTotMarks(99);

        ArrayList<Student> list = new ArrayList<>(); 
        list.add(s4);
        list.add(s3);
        list.add(s1);
        list.add(s2);

        System.out.println("Array before sorting:");
        for(int i=0; i<list.size(); i++){
            System.out.println(list.get(i).toString());
        }

        Student.sortSelection(list);

        System.out.println("Array after sorting:");
        for(int i=0; i<list.size(); i++){
            System.out.println(list.get(i).toString());
        }
    }
}

输出：

Array before sorting:
Name:          Ved Parkash - Total Marks:  99
Name:            Adam Noob - Total Marks: 100
Name:           John Smith - Total Marks:  98
Name:           Jack Smith - Total Marks:  98
Array after sorting:
Name:           Jack Smith - Total Marks:  98
Name:           John Smith - Total Marks:  98
Name:          Ved Parkash - Total Marks:  99
Name:            Adam Noob - Total Marks: 100

笔记：

1)看到添加到列表中的学生的顺序,是4,3,1,然后是2.这是为了证明当年级匹配时,它会根据名称进行排序(杰克·史密斯与约翰·史密斯).

2)我对学生进行了硬编码,以使其更好地进行演示.

3)您可能会注意到,由于问题仅与排序有关,因此我没有设置任何其他变量,并且对排序唯一起作用的变量是：name和totMarks.您将不得不做剩下的事情.

4)我使用的是ArrayList,但不仅限于此,只需进行简单的更改,就可以在普通的Student []数组上使用它.

5)较大的函数不必是静态的,您可以使其成为成员函数并以不同的方式使用它.例如,上面的代码将更改为：

    public boolean larger(Student other){
        if(totMarks < other.totMarks) return false; 
        else if (totMarks > other.totMarks) return true;
        // compare names
        else return name.compareTo(other.name) > 0;
    }

    public static ArrayList<Student> sortSelection(ArrayList<Student> list){
        for(int i=0; i<list.size(); i++){
            for(int j=i+1; j< list.size(); j++){
                // comparison way changes accordingly
                if(list.get(i).larger(list.get(j))){ // swap
                    Student temp = list.get(i); 
                    list.set(i,temp);
                }
            }
        }
        return list;
    }