给多个方法提供一个数组

Question

例如，我们有5门课程: comp2011、comp2012、comp2013、comp2014、comp2015。

在添加或删除课程时，您需要指定学生的名称(您可以假设它是唯一的)和将要添加/删除的课程。

对于添加课程，如果学生已经注册了至少一门课程，系统将将新添加的课程附加到其课程记录中。但是，如果这是学生第一次尝试添加课程，系统将为他/她创建一个记录，并声明这是该学生的第一堂课(见下面的示例输出).

该系统应支持列出学生已注册的课程。

系统应该支持一些基本的验证检查:学生不能添加两次相同的课程，也不能放弃未注册的课程。

当删除最后一门课程时，学生课程列表中将没有任何内容，但是系统仍然应该为他/她保留空文件，以便在添加另一门课程时，它知道这不是第一次。

如何使数组同时支持我的所有方法，知道我已经创建了一个添加、删除、列表和退出的方法。(退出就是没有数组)

这就是我现在的处境：

import java.util.*;
public class codeTraining {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        String AddDrop = "";
        while (!AddDrop.equals("Q")) {
            System.out.println("Add-Drop Course Menu");
            System.out.println("A for Add, D for Drop, L for List, Q for Quit");
            AddDrop = input.nextLine();
            switch (AddDrop) {
                case "A" -> A();
                case "D" -> D();
                case "L" -> L();
                case "Q" -> Q();
                default -> System.out.println("Please enter a valid Character!");
            }
        }
    }

    public static void A() {
        Scanner input = new Scanner(System.in);
        String name;
        String course;
        System.out.println("Please enter the student name:");
        name = input.nextLine();
        System.out.println("Please enter the course you want to add:");
        course = input.nextLine();
        System.out.println("Adding course : " + course);
        System.out.println("Add course successfully");
    }

    public static void D() {
        Scanner input = new Scanner(System.in);
        System.out.println("Please enter a student name");
        String name = input.nextLine();
        System.out.println("please enter the course you want to drop:");
        String course = input.nextLine();
        System.out.println("Dropping course: " + course);
        System.out.println("Drop course successfully");
    }

    public static void L() {
        Scanner input = new Scanner(System.in);
        String name, course;
        System.out.println("Please enter the student name");
        name = input.nextLine();
    }

    public static void Q() {
        System.out.println("Quit...");
    }
}

Answer 1

对于这个问题，您可以使用hashmap，因为它将允许您快速、轻松地查找学生的课程。它被声明为所有方法之外的全局变量，这意味着任何方法都可以访问它。

import java.util.*;

public class codeTraining {
    //Hashmap contains names that are mapped to lists of courses
    static HashMap<String, ArrayList<String>> map =
            new HashMap<String, ArrayList<String>>();

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        String AddDrop = "";
        while (!AddDrop.equals("Q")) {
            System.out.println("Add-Drop Course Menu");
            System.out.println("A for Add, D for Drop, L for List, Q for Quit");
            AddDrop = input.nextLine();
            //added breaks to the switch statement
            switch (AddDrop) {
                case "A": A(); break;
                case "D": D(); break;
                case "L": L(); break;
                case "Q": Q(); break;
                default: System.out.println("Please enter a valid Character!");
            }
        }
    }

    public static void A() {
        Scanner input = new Scanner(System.in);
        String name;
        String course;
        System.out.println("Please enter the student name:");
        name = input.nextLine();
        System.out.println("Please enter the course you want to add:");
        course = input.nextLine();
        System.out.println("Adding course : " + course);
        //if the map contains the name already
        if (map.containsKey(name)) {
            //if the student already had the course in its list
            if (map.get(name).contains(course)) {
                System.out.println("The student is already taking this course");
                return;
            } else {
                //adding the course to the list that's mapped
                //to the student's name in the hashmap 
                map.get(name).add(course);
            }
        }
        //if student isn't in the system
        else {
            //create a new arraylist with the new course in it
            ArrayList<String> courses = new ArrayList<String>();
            courses.add(course);
            //add a map between the student's name and his course list
            map.put(name, courses);
        }
        System.out.println("Add course successfully");
    }

    public static void D() {
        Scanner input = new Scanner(System.in);
        System.out.println("Please enter a student name");
        String name = input.nextLine();
        System.out.println("please enter the course you want to drop:");
        String course = input.nextLine();
        System.out.println("Dropping course: " + course);
        //checks if name in system
        if (map.containsKey(name)) {
            if (map.get(name).contains(course)) {
                //removing if list had the course in it
                map.get(name).remove(course);
            } else {
                System.out.println("This student isn't taking that course");
                return;
            }
        } else {
            System.out.println("This student isn't in the system");
            return;
        }
        System.out.println("Drop course successfully");
    }

    public static void L() {
        Scanner input = new Scanner(System.in);
        String name, course;
        System.out.println("Please enter the student name");
        name = input.nextLine();
        if (map.containsKey(name)) {
            //printing out courses for specified student
            ArrayList<String> courses = map.get(name);
            for (int i = 0; i < courses.size(); i++) {
                System.out.print(courses.get(i) + " ");
            }
        } else {
            System.out.println("This student isn't in the system");
        }
    }

    public static void Q() {
        System.out.println("Quit...");
    }
}

我添加了一些注释，以防您对hashmap不熟悉。基本上，map.put(key, value)将向列表中添加一个映射对，以便当您调用map.get(key)时，它将返回您将其放入的任何值。我希望这能帮上忙，请随便问问你是否感到困惑。