Java基础之年龄、姓名排序
问题：一个ArrayList集合里有很多person类的实例，如何通过年龄和姓名对这些类进行排序。
法一：
通过Collections提供的sort方法进行排序，通过实现Comparator 接口重写compare方法；
1 import java.util.*; 2 3 public class Person { 4 public int age; 5 public String name; 6 7 public Person() { 8 } 9 10 public Person(String name, int age) { 11 this.name = name; 12 this.age = age; 13 } 14 15 16 @Override 17 public String toString() { 18 return "Person{" + 19 "age=" + age + 20 ", name='" + name + '\'' + 21 '}'; 22 } 23 24 public static void main(String[] args) { 25 ArrayList<Person> people = new ArrayList<>(); 26 27 people.add(new Person("1", 18)); 28 people.add(new Person("2", 18)); 29 people.add(new Person("4", 19)); 30 people.add(new Person("3", 19)); 31 people.add(new Person("6", 17)); 32 people.add(new Person("5", 17)); 33 people.add(new Person("8", 20)); 34 people.add(new Person("7", 20)); 35 36 Collections.sort(people, new Comparator<Person>() { 37 @Override 38 public int compare(Person o1, Person o2) { 39 if (o1.age - o2.age != 0) 40 return o2.age - o1.age; 41 return o2.name.compareTo(o1.name); 42 } 43 }); 44 System.out.println(people); 45 46 }
47 }
法二:
对象直接调用sort()方法，用Lambda表达式(其实是和上面的一样的)
1 import java.util.*; 2 3 public class Person { 4 public int age; 5 public String name; 6 7 public Person() { 8 } 9 10 public Person(String name, int age) { 11 this.name = name; 12 this.age = age; 13 } 14 15 @Override 16 public String toString() { 17 return "Person{" + 18 "age=" + age + 19 ", name='" + name + '\'' + 20 '}'; 21 } 22 23 public static void main(String[] args) { 24 ArrayList<Person> people = new ArrayList<>(); 25 26 people.add(new Person("1", 18)); 27 people.add(new Person("2", 18)); 28 people.add(new Person("4", 19)); 29 people.add(new Person("3", 19)); 30 people.add(new Person("6", 17)); 31 people.add(new Person("5", 17)); 32 people.add(new Person("8", 20)); 33 people.add(new Person("7", 20)); 34 35 people.sort(((o1, o2) -> {if (o1.age - o2.age != 0) 36 return o1.age - o2.age; 37 return o1.name.compareTo(o2.name);})); 38 39 System.out.println(people); 40 41 } 42 }
