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

让你的person类实现Comparable< person >,然后实现compareTo方法,例如:

public int compareTo(Person o) {
    int result = name.compareToIgnoreCase(o.name);
    if(result==0) {
        return Integer.valueOf(age).compareTo(o.age);
    }
    else {
        return result;
    }
}

它将首先按名称(大小写不敏感)排序,然后按年龄排序。然后可以在Person对象的集合或数组上运行Arrays.sort()或Collections.sort()。

其他回答

Guava的ComparisonChain提供了一种干净的方法。参考这个链接。

用于执行链式比较语句的实用程序。例如:

   public int compareTo(Foo that) {
     return ComparisonChain.start()
         .compare(this.aString, that.aString)
         .compare(this.anInt, that.anInt)
         .compare(this.anEnum, that.anEnum, Ordering.natural().nullsLast())
         .result();
   }

根据需要创建尽可能多的比较器。之后,为每个订单类别调用“then comparison”方法。这是Streams的一种做法。看到的:

//Sort by first and last name
System.out.println("\n2.Sort list of person objects by firstName then "
                                        + "by lastName then by age");
Comparator<Person> sortByFirstName 
                            = (p, o) -> p.firstName.compareToIgnoreCase(o.firstName);
Comparator<Person> sortByLastName 
                            = (p, o) -> p.lastName.compareToIgnoreCase(o.lastName);
Comparator<Person> sortByAge 
                            = (p, o) -> Integer.compare(p.age,o.age);

//Sort by first Name then Sort by last name then sort by age
personList.stream().sorted(
    sortByFirstName
        .thenComparing(sortByLastName)
        .thenComparing(sortByAge)
     ).forEach(person->
        System.out.println(person));        

在多个字段上对用户定义的对象进行排序- Comparator (lambda stream)

让你的person类实现Comparable< person >,然后实现compareTo方法,例如:

public int compareTo(Person o) {
    int result = name.compareToIgnoreCase(o.name);
    if(result==0) {
        return Integer.valueOf(age).compareTo(o.age);
    }
    else {
        return result;
    }
}

它将首先按名称(大小写不敏感)排序,然后按年龄排序。然后可以在Person对象的集合或数组上运行Arrays.sort()或Collections.sort()。

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。