Java Collections.sort()排序程式碼案例
阿新 • • 發佈:2020-02-20
1、案例:
Person物件(名字,id,年齡)
要求按照,年齡從小到大排序,年齡相等,按照名字的字典順序de倒序排序
2、案例設計:
1)使用ArrayList儲存Person物件,
2)利用Collections.sort()進行排序
3)輸出結果
3、程式碼分享:
package CollectionDemo; import java.util.ArrayList; import java.util.Collections; import java.util.List; import java.util.Comparator; class Person{ private int id; private int age; private String name; public Person(int id,int age,String name){ this.id=id; this.age=age; this.name=name; } public void setId(int id) { this.id = id; } public void setAge(int age) { this.age = age; } public void setName(String name){ this.name=name; } public int getId() { return id; } public int getAge() { return age; } public String getName() { return name; } @Override public String toString() { return "Person{" + "id=" + id + ",age=" + age + ",name='" + name + '\'' + '}'; } } public class CollectionDemo2 { public static void main(String[] args){ List<Person> arrayList =new ArrayList<>(); arrayList.add(new Person(001,20,"yang")); arrayList.add(new Person(002,"zhang")); arrayList.add(new Person(003,30,"li")); arrayList.add(new Person(004,40,"Coco")); arrayList.add(new Person(005,"Marry")); Collections.sort(arrayList,new Comparator<Person>(){ public int compare(Person o1,Person o2){ if(o1.getAge()!=o2.getAge()){ return o1.getAge()-o2.getAge();//按照年齡升序排序 }else{ return o2.getName().compareToIgnoreCase(o1.getName());//按照名字的字典順序倒序排序 } } }); //輸出 for(Person p:arrayList){ System.out.println(p); } } }
輸出:
Person{id=2,age=20,name='zhang'}
Person{id=1,name='yang'}
Person{id=3,age=30,name='li'}
Person{id=5,age=40,name='Marry'}
Person{id=4,name='Coco'}
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