Cry*_*tal 1 java sorting collections
我正在努力熟悉收藏.我有一个String,它是我的密钥,电子邮件地址和Person对象(firstName,lastName,电话,电子邮件).我在Sun的网页上的Java集合章节中读到,如果你有一个HashMap并希望它被排序,你可以使用TreeMap.这种排序如何运作？它是基于Person类中的compareTo()方法吗？我覆盖了Person类中的compareTo()方法以按lastName排序.但它不能正常工作,并且想知道我是否有正确的想法.这段代码底部的getSortedListByLastName是我尝试转换为TreeMap的地方.此外,如果这是正确的方法,或者正确的方法之一,我如何按firstName排序,因为我的compareTo()是通过lastName进行比较.
import java.util.*;
public class OrganizeThis
{
/**
Add a person to the organizer
@param p A person object
*/
public void add(Person p)
{
staff.put(p.getEmail(), p);
//System.out.println("Person " + p + "added");
}
/**
* Remove a Person from the organizer.
*
* @param email The email of the person to be removed.
*/
public void remove(String email)
{
staff.remove(email);
}
/**
* Remove all contacts from the organizer.
*
*/
public void empty()
{
staff.clear();
}
/**
* Find the person stored in the organizer with the email address.
* Note, each person will have a unique email address.
*
* @param email The person email address you are looking for.
*
*/
public Person findByEmail(String email)
{
Person aPerson = staff.get(email);
return aPerson;
}
/**
* Find all persons stored in the organizer with the same last name.
* Note, there can be multiple persons with the same last name.
*
* @param lastName The last name of the persons your are looking for.
*
*/
public Person[] find(String lastName)
{
ArrayList<Person> names = new ArrayList<Person>();
for (Person s : staff.values())
{
if (s.getLastName() == lastName) {
names.add(s);
}
}
// Convert ArrayList back to Array
Person nameArray[] = new Person[names.size()];
names.toArray(nameArray);
return nameArray;
}
/**
* Return all the contact from the orgnizer in
* an array sorted by last name.
*
* @return An array of Person objects.
*
*/
public Person[] getSortedListByLastName()
{
Map<String, Person> sorted = new TreeMap<String, Person>(staff);
ArrayList<Person> sortedArrayList = new ArrayList<Person>();
for (Person s: sorted.values()) {
sortedArrayList.add(s);
}
Person sortedArray[] = new Person[sortedArrayList.size()];
sortedArrayList.toArray(sortedArray);
return sortedArray;
}
private Map<String, Person> staff = new HashMap<String, Person>();
public static void main(String[] args)
{
OrganizeThis testObj = new OrganizeThis();
Person person1 = new Person("J", "W", "111-222-3333", "JW@ucsd.edu");
Person person2 = new Person("K", "W", "345-678-9999", "KW@ucsd.edu");
Person person3 = new Person("Phoebe", "Wang", "322-111-3333", "phoebe@ucsd.edu");
Person person4 = new Person("Nermal", "Johnson", "322-342-5555", "nermal@ucsd.edu");
Person person5 = new Person("Apple", "Banana", "123-456-1111", "apple@ucsd.edu");
testObj.add(person1);
testObj.add(person2);
testObj.add(person3);
testObj.add(person4);
testObj.add(person5);
System.out.println(testObj.findByEmail("JW@ucsd.edu"));
System.out.println("------------" + '\n');
Person a[] = testObj.find("W");
for (Person p : a)
System.out.println(p);
System.out.println("------------" + '\n');
a = testObj.find("W");
for (Person p : a)
System.out.println(p);
System.out.println("SORTED" + '\n');
a = testObj.getSortedListByLastName();
for (Person b : a) {
System.out.println(b);
}
}
}
人员类:
public class Person implements Comparable
{
String firstName;
String lastName;
String telephone;
String email;
public Person()
{
firstName = "";
lastName = "";
telephone = "";
email = "";
}
public Person(String firstName)
{
this.firstName = firstName;
}
public Person(String firstName, String lastName, String telephone, String email)
{
this.firstName = firstName;
this.lastName = lastName;
this.telephone = telephone;
this.email = email;
}
public String getFirstName()
{
return firstName;
}
public void setFirstName(String firstName)
{
this.firstName = firstName;
}
public String getLastName()
{
return lastName;
}
public void setLastName(String lastName)
{
this.lastName = lastName;
}
public String getTelephone()
{
return telephone;
}
public void setTelephone(String telephone)
{
this.telephone = telephone;
}
public String getEmail()
{
return email;
}
public void setEmail(String email)
{
this.email = email;
}
public int compareTo(Object o)
{
String s1 = this.lastName + this.firstName;
String s2 = ((Person) o).lastName + ((Person) o).firstName;
return s1.compareTo(s2);
}
public boolean equals(Object otherObject)
{
// a quick test to see if the objects are identical
if (this == otherObject) {
return true;
}
// must return false if the explicit parameter is null
if (otherObject == null) {
return false;
}
if (!(otherObject instanceof Person)) {
return false;
}
Person other = (Person) otherObject;
return firstName.equals(other.firstName) && lastName.equals(other.lastName) &&
telephone.equals(other.telephone) && email.equals(other.email);
}
public int hashCode()
{
return this.email.toLowerCase().hashCode();
}
public String toString()
{
return getClass().getName() + "[firstName = " + firstName + '\n'
+ "lastName = " + lastName + '\n'
+ "telephone = " + telephone + '\n'
+ "email = " + email + "]";
}
}
实际上,你得到了错误的想法.
这是要点:
Map<K,V>是一个映射K key到V valueTreeMap<K,V>是一个SortedMap<K,V>排序键,而不是值因此,a TreeMap<String,Person>将根据电子邮件地址排序,而不是基于Person姓名.
如果你需要一个SortedSet<Person>,或者一个排序List<Person>然后那是一个不同的概念,是的Person implements Comparable<Person>,或者a Comparator<Person>会派上用场.
java.lang.Comparable<T> - 定义类型对象的"自然排序"java.util.Comparator<T> - 定义类型对象的"自定义"比较java.util.Map<K,V> - 将键映射到值,而不是相反java.util.SortedMap<K,V> - 对键进行排序,而不是值java.util.SortedSet<E> - 订购的套装java.util.Collections.sort(List) - 一种实用的排序方法
Comparator已经有很多例子,但还有一个例子:
import java.util.*;
public class Example {
static String lastName(String fullName) {
return fullName.substring(fullName.indexOf(' ') + 1);
}
public static void main(String[] args) {
Map<String,String> map = new TreeMap<String,String>();
map.put("001", "John Doe");
map.put("666", "Anti Christ");
map.put("007", "James Bond");
System.out.println(map);
// "{001=John Doe, 007=James Bond, 666=Anti Christ}"
// Entries are sorted by keys!
// Now let's make a last name Comparator...
Comparator<String> lastNameComparator = new Comparator<String>() {
@Override public int compare(String fullName1, String fullName2) {
return lastName(fullName1).compareTo(lastName(fullName2));
}
};
// Now let's put all names in a SortedSet...
SortedSet<String> namesByLastName =
new TreeSet<String>(lastNameComparator);
namesByLastName.addAll(map.values());
System.out.println(namesByLastName);
// "[James Bond, Anti Christ, John Doe]"
// Names sorted by last names!
// Now let's use a List instead...
List<String> namesList = new ArrayList<String>();
namesList.addAll(map.values());
System.out.println(namesList);
// "[John Doe, James Bond, Anti Christ]"
// These aren't sorted yet...
Collections.sort(namesList);
System.out.println(namesList);
// "[Anti Christ, James Bond, John Doe]"
// Sorted by natural ordering!
// Now let's sort by string lengths...
Collections.sort(namesList, new Comparator<String>() {
@Override public int compare(String s1, String s2) {
return Integer.valueOf(s1.length()).compareTo(s2.length());
}
});
System.out.println(namesList);
// "[John Doe, James Bond, Anti Christ]"
// SUCCESS!!!
}
}
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