查找分配给ArrayList中的字符串的相同对象

Question

我不确定自己是否正确表达了自己的标题，但我会尽力在此处更具体。

我有什么： 的的ArrayList与客户/顾客。 安排数组列表与电话号码分配给客户/客户。

我的客户端类： import java.util。*;

public class Clients implements Comparable<Clients> { 

    private String name; 
    private String address; 

    public ArrayList<Share> shareList = new ArrayList<Share>(); 

    private PhoneBook phoneBook = new PhoneBook(); 

    public Clients(String name, String address) { 

     this.name = name; 
     this.address = address; 
    } 


    public void addPhoneDescription(String description) { 
     phoneBook.addPhoneDescription(description); 
    } 

    public void addPhoneNumber(String number) { 
     phoneBook.addPhoneNumber(number); 
    } 


    public String getName() { 
     return name; 
    } 

    public void setName(String name) { 
     this.name = name; 
    } 

    public String getAddress() { 
     return address; 
    } 

    public void setAddress(String address) { 
     this.address = address; 
    } 

    public PhoneBook getPhoneBook() { 
     return phoneBook; 
    } 

    public boolean equals(Object obj) { 
     Clients c = (Clients) obj; 
     return this.name.equals(c.name); 
    } 

    public String toString() { 
     String result; 

     result = ("\n" + "Name: " + this.getName() + "\n" + "Address: " 
       + this.getAddress() + "\n" + "Phone description: " + this 
       .getPhoneBook()); 

     return result; 
    } 

    public int compareTo(Clients other) { 
     return name.compareTo(other.name); 

    } 

} 

这是我电话簿类由一套/ get方法：

import java.util.ArrayList; 

public class PhoneBook { 

    private ArrayList<String> numbersList = new ArrayList<String>(); 
    private ArrayList<String> phoneDescription = new ArrayList<String>(); 

    public void addPhoneDescription(String description) { 
     phoneDescription.add(description); 
    } 

    public void addPhoneNumber(String number) { // Add the phone number to the numbersList 

     numbersList.add(number); 

    } 

    public String toString(){ 
     return numbersList.toString() + phoneDescription.toString(); 
    } 


} 

我想实现：如果我创造让说，3个客户端，其中2具有相同的号码我想打印出他们分享/拥有共同号码的2位客户，等等。我有我的计划类中创建一个方法了：

public void findDuplicatedNumbers() { 

     // method that looks for duplicated numbers that clients have 

     ArrayList<Integer> sNumber = new ArrayList<>(); 
     ArrayList<Integer> duplicateNumber = new ArrayList<>(); 

     for (int i = 0; i < clientList.size(); i++) { 
      for (int k = 0; k < (clientList.get(i).getPhoneBook().size); k++) { 
       if (sNumber.contains(clientList.get(i).getPhoneBook().get(k).getNumber())) { 
        if (duplicateNumber.contains(clientList.get(i).getPhoneBook().get(k).getNumber())) { 

       } else { 
         // adds to duplicateNumber arrayList 
         duplicateNumber.add(clientList.get(i).getPhoneBook().get(k).getNumber()); 
        } 
       } else { 
        // adds to sNumber arrayList 
        sNumber.add(clientList.get(i).getPhoneBook().get(k).getNumber()); 
       } 
      } 

     } 
     for (int i = 0; i < duplicateNumber.size(); i++) { 
      System.out.println("Phone number: " + duplicateNumber.get(i) + " is share with these clients: "); 
      for (int k = 0; k < clientList.size(); k++) { 
       for (int p = 0; p < (clientList.get(p).getPhoneBook().size()); p++) { 
        if (duplicateNumber.get(i) == clientList.get(k).getPhoneBook().get(p).getNumber()) { 
         System.out.println(clientList.get(k).getName() + ", "); 
        } 
       } 
      } 
      System.out.println("\n"); 
     } 


    } 

Answer 1

PUH ..安静了一个简单的任务很多代码。 KISS（保持简单）：

public void findDuplicatedNumbers() { 
    Map<String, Set<Client> > duplicates = 
     new HashMap<String, Set<Client> >(); 
    Set<String> phoneNumbers = new HashSet<String>(); 

    for(Client client : clientList) { 
     PhoneBook phoneBook = client.getPhoneBook(); 
     phoneNumbers.addAll(phoneBook.getNumberList()); 
    } 

    for(String phoneNumber : phoneNumbers) { 
     Set<Client> clients = findClientsByPhoneNumber(phoneNumber); 

     if(clients.size() > 1) 
      duplicates.put(phoneNumber, clients); 
    } 

    for(Entry<String, Set<Client> entry : duplicates.entrySet()) { 

      System.out.println("phonenumber " + entry.getKey() + " is dubplicated/is share with these clients:"); 
      for(Client client : entry.getValue()) { 
      System.out.println(client.getName()); 
      } 

    } 
} 
protected Set<Client> findClientsByPhoneNumber(String phoneNumber) { 
    Set<Client> clients = new HashSet<Client>(); 

    for(Client client : clientList) { 
     List<String> phoneNumbers = client.getPhonebook().getNumberList(); 
     if(phoneNumbers.contains(phoneNumber)) { 
      clients.add(client); 
     } 
    } 
    return clients; 
} 

这不仅是清洁的结构，并实现了由寻找客户数量也更多的表现在大多数情况下，额外的接口。 只需清理你想要做的事情：如果你想存储一个唯一对象列表，它就是Set实现之一。如果你想通过键存储某些东西，可以使用Map实现。

Answer 2

public class PhoneBook { 

    private ArrayList<String> numbersList = new ArrayList<String>(); 
    private ArrayList<String> phoneDescription = new ArrayList<String>(); 

    public void addPhoneDescription(String description) { 
     phoneDescription.add(description); 
    } 

    public void addPhoneNumber(String number) { // Add the phone number to the numbersList 

     numbersList.add(number); 

    } 

    public String toString(){ 
     return numbersList.toString() + phoneDescription.toString(); 
    } 

    public ArrayList<String> findDuplicatedNumbers() { 
     ArrayList<String> dupes = new ArrayList<String>(); 
     Collections.sort(numbersList); 
     Iterator<String> iter = numbersList.iterator(); 
     while(iter.hasNext()) { 
      String next = iter.next(); 
      if(iter.hasNext() && next == iter.next()) { 
       dupes.add(next); 
      } 
     } 

     return dupes; 
    } 
}