Java地址簿。如何防止代码中重复的联系人？

Question

switch(menuChoice) {

case 1: 
    System.out.println("Enter your contact's first name:\n");
    String fname = scnr.next();
    System.out.println("Enter your contact's last name:\n");
    String lname = scnr.next();
    Necronomicon.addContact(new Person(fname, lname));
    break;

// main truncated here for readability

import java.util.ArrayList;

public class AddressBook {

   ArrayList<Person> ArrayOfContacts= new ArrayList<Person>();

   public void addContact(Person p) {
      ArrayOfContacts.add(p);

 /* 
    for(int i = 0;  i < ArrayOfContacts.size(); i++) {
       if(ArrayOfContacts.get(i).getID() != p.getID())
           ArrayOfContacts.add(p);

    else 
       System.out.println("Sorry this contact already exists.");
    }       
  */

    }
}

public class Person {

   private String fName = null;
   private String lName = null;
   private static int ID = 1000;

   public Person(String fName, String lName) {       // Constructor I'm using to try and increment the ID each time a Person object is created starting at 1001.

     this.fName = fName;
     this.lName = lName;
     ID = ID + 1;
   }
}

我正在尝试创建一个通讯录，其中每个联系人都有一个名字，姓氏和唯一的ID。

我的问题是如何防止用户输入具有相同名字和姓氏的重复联系人？我应该在addContact方法中还是在main中实现某种检查？怎么样？

Answer 1

这是用于保留重复ID的代码。

public void addContact(Person p) {

    for(int i = 0;  i < ArrayOfContacts.size(); i++) {
        Person contact = ArrayOfContacts.get(i);
        if(contact.getID() == p.getID()) {
            System.out.println("Sorry this contact already exists.");
            return; // the id exists, so we exit the method. 
        }
    }

    // Otherwise... you've checked all the elements, and have not found a duplicate
    ArrayOfContacts.add(p);

}

如果您想更改此代码以保留重复的名称，请执行以下操作

public void addContact(Person p) {
    String pName = p.getFname() + p.getLname();
    for(int i = 0;  i < ArrayOfContacts.size(); i++) {
        Person contact = ArrayOfContacts.get(i);
        String contactName =  contact.getFname() + contact.getLname(); 
        if(contactName.equals(pName)) {
            System.out.println("Sorry this contact already exists.");
            return; // the name exists, so we exit the method. 
        }
    }

    // Otherwise... you've checked all the elements, and have not found a duplicate
    ArrayOfContacts.add(p);

}