查找分配给 ArrayList 中的字符串的相同对象

Question

我不确定我是否正确地表达了自己的标题，但我会尝试在这里更具体。

我有什么：具有客户端/客户的数组列表。一个数组列表，其中包含分配给客户/客户的电话号码。

我的客户端类： 导入 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);
    }
}

这是我的电话簿类，由set/get方法组成：

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();
    }

}

我想要实现的目标：因此，如果我创建假设 3 个客户端并且其中 2 个具有相同的数字，我想打印出这 2 个客户端以及它们共享/具有的共同编号，依此类推。我在程序类上创建了一个方法：

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 .. 为一个简单的任务安静了很多代码。亲吻（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;
}

这不仅结构更简洁，并实现了用于按编号查找客户端的附加接口，而且在大多数情况下具有更高的性能。只要明确你想要做什么：如果你想存储一个唯一对象的列表，它是Set - 实现之一。如果要通过键存储某些内容，请使用 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;
    }
}