java迭代器在找到匹配项后中途停止

Question

//Example from class that I was trying to base off of:

public void printMajors(String major) {
for(Student x : dir) {

//dir = new HashMap<Integer, Student>(); and Student is the second class.

if (x.getMajor().equals(major)) {
System.out.println(x.getName());
    }
}

}

我想打印出所有插入了相同区号的人，但迭代器在找到匹配项后中途停止。我如何让它继续循环并打印出每个人的区号相同

//代码中的问题

public boolean printAreaCode(String areacode) {
    for (String p : phonebook.values()) {
        if (p.startsWith(areacode)) {
            for (String name : phonebook.keySet()) {
                String key = name;
                System.out.println(key);
                return true;
            }
        }
    }
    return false;
}

整个任务

import java.util.HashMap;
import java.util.Iterator;

public class PhoneBook {

    private HashMap<String, String> phonebook;

    public static void main(String[] args) {
    PhoneBook phone = new PhoneBook();
    phone.addEntry("Alexander Schnell", "987-654-3210");
    phone.addEntry("Bob the Builder", "555-555-5555");
    phone.addEntry("Michael", "465-858-5555");
    phone.addEntry("Robert", "778-555-1234");
    phone.addEntry("Charlie", "987-546-4564");
    phone.addEntry("Steve", "909-555-7845");
    System.out.println("All of the people in the directory:");
    phone.printListings(); //works
    System.out.println("");
    System.out.println("People with matching area codes:"); //WIP
    phone.printAreaCode("987");
    System.out.println("");
    System.out.println("Prints phone number for requested person:");
    phone.getNumber("Bob the Builder"); //works
}
//Adds a new entry to the phone book (naturally).
//It has two parameters, both Strings: the person’s name and their phone
//number (in the form “610-499-4035”).

public void addEntry(String name, String number) {
    phonebook.put(name, number);
}
//prints the names and phone numbers of everyone in the phone book

private void printListings() {
    for (Iterator<String> it = phonebook.keySet().iterator(); it.hasNext();) {
        for (String p : phonebook.values()) {
            String key = (String) it.next();
            System.out.println(key + ": " + p.toString());
        }
    }
}
//looks up a number in the map.
//It has one parameter (a person’s name) and returns a PhoneNumber object.

public String getNumber(String name) {
    //public void getNumber(String name) {
    // no return statement
    System.out.println(phonebook.get(name));
    return phonebook.get(name);
}
//prints the names of all people with the given area code

public boolean printAreaCode(String areacode) {
    for (String p : phonebook.values()) {
        if (p.startsWith(areacode)) {
            for (String name : phonebook.keySet()) {
                String key = name;
                System.out.println(key);
                return true;
            }
        }
    }
    return false;
    }

    public PhoneBook() {
        phonebook = new HashMap<String, String>();
    }
}

我甚至不认为这门课能和我的主课交流。 我的老师刚才说必须包括在内。 这可能是个问题

public class PhoneNumber {
    private String areacode;
    private String prefix;
    private String four;

    //Insert a phone number
    public PhoneNumber(String number) {
        String[] parts = number.split("-");
        areacode = parts[0];
        prefix = parts[1];
        four = parts[2];
    }

    @Override
    public String toString() {
        String s = areacode + "-" + prefix + "-" + four;
        return s;
    }
}

Answer 1

您在地图中的迭代错误您的return true;语句是问题的原因，但我们暂时忽略这一点

总是像这样迭代地图：

for(Map.Entry<String, String> entry: phoneBook.entrySet())
{
    if(entry.getValue().startsWith(areaCode))
    {
        System.out.println(entry.getKey());
    }
} 

Answer 2

所以，你需要打印key的value的startsWith{}，所以你的代码基本上是向后的

目前，您可以通过代码扫描来寻找匹配的值，这是可以的，但它会打印所有key值吗

相反，您需要从每个key开始，获取value，然后测试与键关联的value是否与areacode匹配

public boolean printAreaCode(String areacode) {
    boolean found = false;
    for (String key : phonebook.keySet()) {
        String value = phonebook.get(key);
        if (value.startsWith(areacode)) {
            System.out.println(key);
            found = true;
        }
    }
    return found;
}

现在，这是非常低效的。基本上，对于每个迭代，您都要求Map查找与key关联的值，这需要Map搜索它的所有不同条目/节点，试图找到匹配的键。虽然在你的例子中，这可能不是一个巨大的延迟，但确实应该尽可能避免

相反，Map提供了对Entry的访问，它将key和value配对成一个对象，使其访问速度更快，例如

public boolean printAreaCode(String areacode) {
    boolean found = false;
    for (Map.Entry<String, String> entry : phonebook.entrySet()) {
        String value = entry.getValue();
        if (value.startsWith(areacode)) {
            String key = entry.getKey();
            System.out.println(key);
            found = true;
        }
    }
    return found;
}

现在，HashMap是Map的一个实现，所以通常我更喜欢使用

private Map<String, String> phonebook;

而不是

private HashMap<String, String> phonebook;

除非我对某些只在HashMap中定义的功能有特殊需求，否则may code不会在意我以后是否使用TreeMap

Answer 3

你不能加这一行import java.util.Map;

我很好奇为什么你不能添加import语句；这是标准的一部分还是你不知道怎么做

你必须在地图上迭代两次。是的，这是非常低效的代码

public boolean printAreaCode(String areacode) {
boolean haveMatch = false;
for (String p : phonebook.values()) {
    if (p.startsWith(areacode)) {
        String value = p;
        for (String name : phonebook.keySet()) {
            if(value.equals(phonebook.get(name)) {
                System.out.println(key);
                haveMatch = true;
        }
    }
}
return haveMatch;

}

Answer 4

我相信你的目的是打印电话号码，其中代码以某个值开头。因此，我们不必从该方法返回任何内容，您可以将返回类型设置为void。你可以让它类似于你的printListings

public void printAreaCode(String areacode) {
    for (String p : phonebook.values()) {
        if (p.startsWith(areacode)) {
            for (String name : phonebook.keySet()) {
                String key = name;
                System.out.println(key);

            }
        }
    }

    }