Question

我不确定我是否正确地表达了自己的标题，但我会在这里尝试更具体。

我有什么： 客户/客户 arraylist 。将电话号码分配给客户/客户的 arraylist 。

我的客户类： 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); } }

这是我的 PhoneBook类，包含set / 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(); } }

我想要实现的目标：所以，如果我创建让我们说3个客户端，其中2个具有相同的数字，我想打印出这两个客户端共享/共享的数量，依此类推。我在程序类中创建了一个方法：

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 ..为一项简单的任务安静了很多代码。 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;
}

这不仅结构清晰，而且还实现了一个额外的界面，可以按编号查找客户端，但在大多数情况下还可以提高性能。只是要清楚你想要做什么：如果你想存储一个唯一对象列表，它就是Set - implementation之一。如果您想通过密钥存储某些内容，请使用Map - implementation。

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;
    }
}

查找分配给ArrayList中的字符串的相同对象

2 个答案: