【发布时间】：2015-07-27 02:53:06
【问题描述】：
我正在练习数组排序，我已经成功排序了一个字符串数组。
我的小程序允许用户输入第一个学生人数，然后是每个学生的姓名，最后是每个学生的年级。
但我还想对 int studentGrade 数组进行排序，以便打印输出中的成绩与学生匹配。在这里，我真的被困住了。方法中的更多解释请参见下文：public void sortingAlgorithm
package assignment8exam;
import java.util.Scanner;
import java.util.Arrays;
/**
*
* @author Anders
*/
public class Course {
Scanner sc = new Scanner(System.in);
public void MainMenu() {
System.out.println("Enter data about a student, start by entering how many");
int numbers = sc.nextInt();// amount of student
String studentNames[] = new String[numbers];
int studentGrade[] = new int[numbers];
for (int i = 0; i < numbers; i++) {
System.out.println("Enter name of student");
Scanner name = new Scanner(System.in);
String names = name.nextLine();
studentNames[i] = names;
}
for (int j = 0; j < numbers; j++) {
System.out.println("Enter grade of student");
Scanner gradeSc = new Scanner(System.in);
int grade = gradeSc.nextInt();
studentGrade[j] = grade;
}
sortingArray(studentNames);
System.out.println("------------------------------------\n");
sortAlgorithm(studentNames, studentGrade);
System.out.println("What do you want");
System.out.println("Exit application 1");
System.out.println("Print out all names of the students 2");
System.out.println("Print out all the grades of the students 3");
System.out.println("Print out pairs consisting of “namegrade 4");
System.out.println("Search for a student - 5");
Scanner choice = new Scanner(System.in);
int order = choice.nextInt();
switch (order) {
case 1:
System.exit(1);
case 2:
PrintOutNames(numbers, studentNames);
case 3:
PrintOutGrades(numbers, studentGrade);
case 4:
PrintOutAll(numbers, studentGrade, studentNames);
case 5:
search(numbers, studentGrade, studentNames);
}
}
public static void PrintOutNames(int numbers, String studentNames[]) {
for (int i = 0; i < numbers; i++) {
System.out.println(studentNames[i]);
}
}
public static void PrintOutGrades(int numbers, int studentGrade[]) {
for (int i = 0; i < numbers; i++) {
System.out.println(studentGrade[i]);
}
}
public static void PrintOutAll(int numbers, int studentGrade[], String studentNames[]) {
System.out.println("--------------------------------------------------------\n");
for (int i = 0; i < numbers; i++) {
System.out.println("Name----> " + studentNames[i] + " grade ---> " + studentGrade[i]);
}
}
public static void search(int numbers, int studentGrade[], String studentNames[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter name on student you want to search on ");
String search = sc.nextLine();
for (int i = 0; i < numbers; i++) {
if (search.equals(studentNames[i])) {
System.out.println("Yes we have a student named " + studentNames[i] + " with the Grade " + studentGrade[i] + " \n ");
}
}
}
public static void sortingArray(String studentNames[]) {
Arrays.sort(studentNames);
System.out.println("-------------\n" + Arrays.toString(studentNames));
}
public static void sortAlgorithm(String studentNames[], int studentGrade[]) {
boolean flag = true;
while (flag) {
flag = false;
for (int j = 0; j < studentNames.length - 1; j++) {
for (int i = j + 1; i < studentNames.length; i++) {
if (studentNames[i].compareTo(studentNames[j]) < 0) {
String temp = studentNames[j];
studentNames[j] = studentNames[i];
studentNames[i] = temp;
// Here i want to place another array that sorts the grade?? how do i do that?
}
}
System.out.println(Arrays.toString(studentNames));
System.out.println(Arrays.toString(studentGrade));
}
}
}
}
【问题讨论】：
-
我的第一条建议是避免从多个数组开始 - 拥有一个
Student对象集合，其中每个学生都有一个姓名和一个年级。这将使您的生活更轻松。 （我还阅读了 Java 命名约定...） -
创建一个类
Student，该类使用名称和等级创建。将所有Student对象存储在列表中。然后独立于您对列表进行排序的标准（使用Comparator），您可以访问相关的其他值。 -
有一个学生抽象是关键。
-
有点跑题了，但是你的
switch-case没有breaks