我有对象数组person (int age;字符串名称;)。
我如何按名字的字母顺序排序这个数组，然后按年龄?
你会用哪种算法呢?
当前回答
对于像这样的一本书:
package books;
public class Book {
private Integer id;
private Integer number;
private String name;
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public Integer getNumber() {
return number;
}
public void setNumber(Integer number) {
this.number = number;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "book{" +
"id=" + id +
", number=" + number +
", name='" + name + '\'' + '\n' +
'}';
}
}
用模拟对象对主类进行排序
package books;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class Main {
public static void main(String[] args) {
System.out.println("Hello World!");
Book b = new Book();
Book c = new Book();
Book d = new Book();
Book e = new Book();
Book f = new Book();
Book g = new Book();
Book g1 = new Book();
Book g2 = new Book();
Book g3 = new Book();
Book g4 = new Book();
b.setId(1);
b.setNumber(12);
b.setName("gk");
c.setId(2);
c.setNumber(12);
c.setName("gk");
d.setId(2);
d.setNumber(13);
d.setName("maths");
e.setId(3);
e.setNumber(3);
e.setName("geometry");
f.setId(3);
f.setNumber(34);
b.setName("gk");
g.setId(3);
g.setNumber(11);
g.setName("gk");
g1.setId(3);
g1.setNumber(88);
g1.setName("gk");
g2.setId(3);
g2.setNumber(91);
g2.setName("gk");
g3.setId(3);
g3.setNumber(101);
g3.setName("gk");
g4.setId(3);
g4.setNumber(4);
g4.setName("gk");
List<Book> allBooks = new ArrayList<Book>();
allBooks.add(b);
allBooks.add(c);
allBooks.add(d);
allBooks.add(e);
allBooks.add(f);
allBooks.add(g);
allBooks.add(g1);
allBooks.add(g2);
allBooks.add(g3);
allBooks.add(g4);
System.out.println(allBooks.size());
Collections.sort(allBooks, new Comparator<Book>() {
@Override
public int compare(Book t, Book t1) {
int a = t.getId()- t1.getId();
if(a == 0){
int a1 = t.getNumber() - t1.getNumber();
return a1;
}
else
return a;
}
});
System.out.println(allBooks);
}
}
其他回答
对于那些能够使用Java 8流API的人来说，这里有一个更整洁的方法: lambda和排序
我正在寻找相当于c# LINQ:
.ThenBy(...)
我在Comparator上找到了Java 8的机制:
.thenComparing(...)
下面是演示算法的代码片段。
Comparator<Person> comparator = Comparator.comparing(person -> person.name);
comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));
请查看上面的链接，以获得更简洁的方法，并解释Java的类型推断如何使其与LINQ相比定义起来更笨拙。
下面是完整的单元测试供参考:
@Test
public void testChainedSorting()
{
// Create the collection of people:
ArrayList<Person> people = new ArrayList<>();
people.add(new Person("Dan", 4));
people.add(new Person("Andi", 2));
people.add(new Person("Bob", 42));
people.add(new Person("Debby", 3));
people.add(new Person("Bob", 72));
people.add(new Person("Barry", 20));
people.add(new Person("Cathy", 40));
people.add(new Person("Bob", 40));
people.add(new Person("Barry", 50));
// Define chained comparators:
// Great article explaining this and how to make it even neater:
// http://blog.jooq.org/2014/01/31/java-8-friday-goodies-lambdas-and-sorting/
Comparator<Person> comparator = Comparator.comparing(person -> person.name);
comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));
// Sort the stream:
Stream<Person> personStream = people.stream().sorted(comparator);
// Make sure that the output is as expected:
List<Person> sortedPeople = personStream.collect(Collectors.toList());
Assert.assertEquals("Andi", sortedPeople.get(0).name); Assert.assertEquals(2, sortedPeople.get(0).age);
Assert.assertEquals("Barry", sortedPeople.get(1).name); Assert.assertEquals(20, sortedPeople.get(1).age);
Assert.assertEquals("Barry", sortedPeople.get(2).name); Assert.assertEquals(50, sortedPeople.get(2).age);
Assert.assertEquals("Bob", sortedPeople.get(3).name); Assert.assertEquals(40, sortedPeople.get(3).age);
Assert.assertEquals("Bob", sortedPeople.get(4).name); Assert.assertEquals(42, sortedPeople.get(4).age);
Assert.assertEquals("Bob", sortedPeople.get(5).name); Assert.assertEquals(72, sortedPeople.get(5).age);
Assert.assertEquals("Cathy", sortedPeople.get(6).name); Assert.assertEquals(40, sortedPeople.get(6).age);
Assert.assertEquals("Dan", sortedPeople.get(7).name); Assert.assertEquals(4, sortedPeople.get(7).age);
Assert.assertEquals("Debby", sortedPeople.get(8).name); Assert.assertEquals(3, sortedPeople.get(8).age);
// Andi : 2
// Barry : 20
// Barry : 50
// Bob : 40
// Bob : 42
// Bob : 72
// Cathy : 40
// Dan : 4
// Debby : 3
}
/**
* A person in our system.
*/
public static class Person
{
/**
* Creates a new person.
* @param name The name of the person.
* @param age The age of the person.
*/
public Person(String name, int age)
{
this.age = age;
this.name = name;
}
/**
* The name of the person.
*/
public String name;
/**
* The age of the person.
*/
public int age;
@Override
public String toString()
{
if (name == null) return super.toString();
else return String.format("%s : %d", this.name, this.age);
}
}
或者，您可以利用Collections.sort()(或Arrays.sort())是稳定的(它不会对相等的元素重新排序)这一事实，并首先使用一个Comparator按年龄排序，然后使用另一个Comparator按名称排序。
在这种特定的情况下，这不是一个很好的主意，但如果你必须能够在运行时改变排序顺序，它可能是有用的。
你可以这样做:
List<User> users = Lists.newArrayList(
new User("Pedro", 12),
new User("Maria", 10),
new User("Rafael",12)
);
users.sort(
Comparator.comparing(User::getName).thenComparing(User::getAge)
);
对于像这样的一本书:
package books;
public class Book {
private Integer id;
private Integer number;
private String name;
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public Integer getNumber() {
return number;
}
public void setNumber(Integer number) {
this.number = number;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "book{" +
"id=" + id +
", number=" + number +
", name='" + name + '\'' + '\n' +
'}';
}
}
用模拟对象对主类进行排序
package books;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class Main {
public static void main(String[] args) {
System.out.println("Hello World!");
Book b = new Book();
Book c = new Book();
Book d = new Book();
Book e = new Book();
Book f = new Book();
Book g = new Book();
Book g1 = new Book();
Book g2 = new Book();
Book g3 = new Book();
Book g4 = new Book();
b.setId(1);
b.setNumber(12);
b.setName("gk");
c.setId(2);
c.setNumber(12);
c.setName("gk");
d.setId(2);
d.setNumber(13);
d.setName("maths");
e.setId(3);
e.setNumber(3);
e.setName("geometry");
f.setId(3);
f.setNumber(34);
b.setName("gk");
g.setId(3);
g.setNumber(11);
g.setName("gk");
g1.setId(3);
g1.setNumber(88);
g1.setName("gk");
g2.setId(3);
g2.setNumber(91);
g2.setName("gk");
g3.setId(3);
g3.setNumber(101);
g3.setName("gk");
g4.setId(3);
g4.setNumber(4);
g4.setName("gk");
List<Book> allBooks = new ArrayList<Book>();
allBooks.add(b);
allBooks.add(c);
allBooks.add(d);
allBooks.add(e);
allBooks.add(f);
allBooks.add(g);
allBooks.add(g1);
allBooks.add(g2);
allBooks.add(g3);
allBooks.add(g4);
System.out.println(allBooks.size());
Collections.sort(allBooks, new Comparator<Book>() {
@Override
public int compare(Book t, Book t1) {
int a = t.getId()- t1.getId();
if(a == 0){
int a1 = t.getNumber() - t1.getNumber();
return a1;
}
else
return a;
}
});
System.out.println(allBooks);
}
}
您需要实现自己的Comparator，然后使用它:例如
Arrays.sort(persons, new PersonComparator());
你的比较器可以看起来像这样:
public class PersonComparator implements Comparator<? extends Person> {
public int compare(Person p1, Person p2) {
int nameCompare = p1.name.compareToIgnoreCase(p2.name);
if (nameCompare != 0) {
return nameCompare;
} else {
return Integer.valueOf(p1.age).compareTo(Integer.valueOf(p2.age));
}
}
}
比较程序首先比较两个名字，如果它们不相等，就返回比较结果，否则就返回比较两人年龄时的比较结果。
这段代码只是一个草稿:因为这个类是不可变的，你可以考虑为它构建一个单例，而不是为每次排序创建一个新实例。
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