问题
我有对象数组person (int age; String name;)。
如何按名称和年龄按字母顺序对此数组进行排序？
你会使用哪种算法？
#1 热门回答(161 赞)
你可以使用Collections.sort如下：
private static void order(List persons) {
Collections.sort(persons, new Comparator() {
public int compare(Object o1, Object o2) {
String x1 = ((Person) o1).getName();
String x2 = ((Person) o2).getName();
int sComp = x1.compareTo(x2);
if (sComp != 0) {
return sComp;
}
Integer x1 = ((Person) o1).getAge();
Integer x2 = ((Person) o2).getAge();
return x1.compareTo(x2);
}});
}
List现在按名称排序，然后按年龄排序。
String.compareTo"按字典顺序比较两个字符串" - 来自docs。
Collections.sort是本机Collections库中的静态方法。它进行实际的排序，你只需要提供一个Comparator来定义你的列表中的两个元素应该如何比较：这是通过提供你自己的compare方法的实现来实现的。
#2 热门回答(86 赞)
对于那些能够使用Java 8流API的人来说，这里有一个更为简洁的方法：Lambdas and sorting
我正在寻找相当于C#LINQ：
.ThenBy(...)
我在Comparator上找到了Java 8中的机制：
.thenComparing(...)
所以这里是演示算法的片段。
Comparator comparator = Comparator.comparing(person -> person.name);
comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));
查看上面的链接，了解更简洁的方法以及有关Java类型推断如何使其与LINQ相比更加笨拙的解释。
以下是完整的单元测试供参考：
@Test
public void testChainedSorting()
{
// Create the collection of people:
ArrayList 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 comparator = Comparator.comparing(person -> person.name);
comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));
// Sort the stream:
Stream personStream = people.stream().sorted(comparator);
// Make sure that the output is as expected:
List 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);
}
}
#3 热门回答(55 赞)
使用Java 8 Streams方法......
//Creates and sorts a stream (does not sort the original list)
persons.stream().sorted(Comparator.comparing(Person::getName).thenComparing(Person::getAge));
而Java 8 Lambda方法......
//Sorts the original list Lambda style
persons.sort((p1, p2) -> {
if (p1.getName().compareTo(p2.getName()) == 0) {
return p1.getAge().compareTo(p2.getAge());
} else {
return p1.getName().compareTo(p2.getName());
}
});
最后...
//This is similar SYNTAX to the Streams above, but it sorts the original list!!
persons.sort(Comparator.comparing(Person::getName).thenComparing(Person::getAge));