java - 使用哈希表的电话簿

Question

我需要帮助创建一个使用哈希表的小型电话簿应用程序。这是我的学校作业。我对 Java 很陌生，所以我无法真正理解这一点。

我对应用程序如何运行有一个基本的代码布局，但我只是不知道如何实现哈希表。

我不允许使用Java中内置的数据结构，所以我必须从头开始构建哈希表。不过，我可以使用哈希表的 native hashCode() 函数。

这是我的代码和其中的一些注释:

import java.util.Scanner;

public class PhoneBook {

public static void main(String[] args) {

    boolean exitPhoneBook = false;
    Scanner userInput = new Scanner(System.in);

    while (exitPhoneBook == false) {

        System.out.println("What do you want to do?");
        System.out.println("1. Add a contact");
        System.out.println("2. Show a contact");
        System.out.println("3. Delete a contact");
        System.out.println("4. Show all contacts");
        System.out.println("5. Exit");
        System.out.print("Select a number: ");

        int action = userInput.nextInt();

        switch (action){
        case 1:
            addContact();
            break;

        case 2:
            showContact();
            break;

        case 3:
            deleteContact();
            break;

        case 4:
            showAll();
            break;

        case 5:
            System.out.println("Goodbye!");
            exitPhoneBook = true;
            break;

        default:
            System.out.println("Invalid option.");
            System.out.print("Select a number: ");
            break;
        }
    }

}

static void addContact(){
    //takes in four strings from user (first name, last name, phone number, email)
}

static void showContact(){
    //takes in two strings from user (first name, last name)
}

static void deleteContact(){
    //takes in two strings from user (first name, last name)
}

static void showAll(){
    //prints out all the contact in the hash table
}

}

Answer 1

对于将其发布为答案，我深表歉意。我不知道如何添加带有添加代码的评论。

编辑:我找出了多值。我现在的问题是删除具有相同索引/散列键的条目。例如。有三个值具有相同的键，我可以删除第一个和第二个值，但不能删除第三个。

这是我的代码:

public void deleteContact(String key) {
    int location = hashFunction(key);
    if (contactsArray[location] == null) { //if the contact doesn't exist
        System.out.println("Contact not found.\n");
        return;
    }

    if (contactsArray[location].key.equals(key)) {
        contactsArray[location] = contactsArray[location].next; //if contact is on first item
        System.out.println("Contact has been removed\n");
        return;
    }

    //If contact is not the first item of the same key
    ContactList prev = contactsArray[location];
    ContactList curr = prev.next;
    while (curr != null && ! curr.key.equals(key)) {
        curr = curr.next;
        prev = curr;
    }

    if (curr != null) {
        prev.next = curr.next;
        System.out.println("Contact has been removed");
        return;
    }
}