3 回答

TA贡献1878条经验 获得超4个赞
由于您不能使用现有的 Comparator 或排序算法,因此您需要自己完成。我已经实现了一个static函数lessOrEqual,它接受 2 个Student实例,比较它们并返回是否s1小于或等于s2。 larger(Student s1, Student s2)仅当s1大于时返回真s2。可以有很多不同的方法来做到这一点,这真的取决于你,因为它只是一个比较。该函数首先检查成绩,如果成绩匹配,它将检查姓名并相应地返回。
编辑:如您所见,我替换lessOrEqual为larger因为我正在使用的选择排序需要找到larger。这是相同的效果,我这样做只是为了更好的可读性。
然后我实现了另一个static接受ArrayList<Student>、排序并返回排序的函数。使用的排序算法非常基本:选择排序。它的 O(N^2) 效率不高,但为了简单起见,我在下面的演示中这样做了。
代码:
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) 查看学生添加到列表中的顺序,它是 4,3, 1 然后是 2。这是为了证明它在成绩匹配时根据姓名排序(Jack Smith vs John Smith)。
2) 我对学生进行硬编码以制作更好的演示。
3) 正如您可能注意到的那样,我没有设置任何其他变量,因为问题完全是关于排序的,唯一对排序有贡献的变量是:name和totMarks。你将不得不做剩下的事情。
4) 我正在使用ArrayList,但不限于此,通过简单的更改,您可以在普通Student[]数组上使用它。
5)函数larger不一定是static,你可以把它做成一个成员函数,用不同的方式使用。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, list.get(j));
list.set(j, temp);
}
}
}
return list;
}

TA贡献1804条经验 获得超8个赞
为了保持简单(即 KISS 原则)并解释我与复合键相关的“提示”,以下是工作示例。
解决方案的“关键”是让树自然地对数据进行排序(不是,恕我直言,通过提供手动排序来添加代码使其变得更加复杂)。因此,学生类需要返回一个树可以自然排序的键。
为了产生所需的排序结果,树的键是(总分,学生姓名)。
这是修改后的 Student 类(减去 getter 和 setter,但我确实添加了一个新的构造函数以使我的生活更轻松):
public class Student {
private String name;
private Integer rollNumber;
private int m1;
private int m2;
private int m3;
private int totMarks;
//Getter setter
public Student() {
}
public Student(String name, Integer rollNumber, int m1, int m2, int m3) {
this.name = name;
this.rollNumber = rollNumber;
this.m1 = m1;
this.m2 = m2;
this.m3 = m3;
this.totMarks = m1 + m2 + m3;
}
public String getKey() {
// return String.format("%d-%s", totMarks, name); // WRONG!
return String.format("%04d-%s", totMarks, name); // Right
}
public String toString() {
return String.format("%06d: %s - %d", rollNumber, name, totMarks);
}
}
请注意,方法中有一行注释掉的代码getKey带有注释WRONG。这与我用个位数分数测试的暗示有关。尝试交换两行代码以查看正确和错误的结果。
这是主要的,我删除了所有扫描仪的东西——再次让我的生活更轻松。希望您可以关注它并重新添加到您的扫描仪循环中。
import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
public class StudentData {
public static void main(String[] args) {
// Initialise a list of students (as I don't want to rekey all the details each
// time I run the program).
List<Student> studentList = Arrays.asList(
new Student("Fred", 1, 2, 2, 2), /* Score of 6 */
new Student("John", 2, 2, 3, 3), /* Score of 8 */
new Student("Jane", 3, 20, 25, 30), /* Score of 75 */
new Student("Julie", 4, 20, 15, 10) /* Score of 45 */
// add as many new students as you like, and reorder them
// as much as you like to see if there is any difference in the
// result (there shouldn't be).
);
// Note the key of the tree is of type String - not Integer.
// This is the heart of the algorithm, the tree will be "sorted"
// on the natural key provided. This "natural key" is in fact
// a compound key that is generated by combining the total score
// and the student name.
Map<String,Student> map = new TreeMap<String,Student>();
for (Student ss : studentList) {
map.put(ss.getKey(),ss);
}
//stdList.forEach(System.out::print);
for(Map.Entry<String,Student> m :map.entrySet()) {
System.out.println(m);
}
}
}
我希望您同意这是一个更简单的解决方案。还有一个潜在的性能优势,因为学生在被加载到树中时被排序(即一次)。我认为这种排序的性能是 log(n)。检索排序可能是 n log(n) 或更糟。

TA贡献1875条经验 获得超3个赞
不是将值存储为 student,而是将它们存储为 (name, student) 的映射,这样当遇到具有相同标记的学生时,就会将其添加到映射中。
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enetr the number of Student");
int totalStudent = sc.nextInt();
Map<Integer, Map<String, Student>> map = new TreeMap<>();
for(int i =0;i<totalStudent;i++) {
Student ss = new Student();
System.out.println("Enter the Student roll number");
ss.setRollNumber(sc.nextInt());
System.out.println("Enter the Student Name");
ss.setName(sc.next());
System.out.println("Enter the m1 marks ");
ss.setM1(sc.nextInt());
System.out.println("Enetr the m2 marks ");
ss.setM2(sc.nextInt());
System.out.println("Enter the m3 marks ");
ss.setM3(sc.nextInt());
ss.setTotMarks(ss.getM1()+ss.getM2()+ss.getM3());
Integer key = ss.getTotMarks();
if (map.get(key) == null){ // if this is a new key in the map, then create a new TreeMap and put it
final TreeMap<String, Student> nameAndStudentMap = new TreeMap<>();
nameAndStudentMap.put(ss.getName(), ss);
map.put(key, nameAndStudentMap);
}else { // if the key already existed, then get the map stored and add to it.
map.get(key).put(ss.getName(), ss);
}
}
//stdList.forEach(System.out::print);
for(Map.Entry<Integer,Map<String, Student>> m :map.entrySet()) {
for (Map.Entry<String, Student> nameAndStudent : m.getValue().entrySet()) {
System.out.println(m.getKey() + " = " + nameAndStudent.getValue());
}
}
}
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