List集合實現對對象引用數據類型的排序

有兩種方式，第一種方式在對象中實現Compartable接口，第二種方式爲創建一個實現實現Comparator接口，或者通過匿名內部類的方式

第一種方式：在對象中實現Compatable接口

//首先創建一個Student對象

package ListSortDemo;

public class Student implements Comparable<Student>{
    private String name;
    private int age;
    private int secore;

    public Student() {
    }

    public Student(String name, int age, int secore) {
        this.name = name;
        this.age = age;
        this.secore = secore;
    }

    @Override
    public String toString() {
        return "Student{" +
                "name='" + name + '\'' +
                ", age=" + age +
                ", secore=" + secore +
                '}'+'\n';
    }


    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public int getSecore() {
        return secore;
    }

    public void setSecore(int secore) {
        this.secore = secore;
    }

    //重寫compareTo()方法
    @Override
    public int compareTo(Student o) {
        return this.secore - o.secore;  //此處通過分數升序排序
    }
}

//測試類
public class ArrayLisetSort {
    public static void main(String[] args) {
         ArrayList<Student> arr = new ArrayList<>();
        arr.add(new Student("張三",18,56));
        arr.add(new Student("李四",15,57));
        arr.add(new Student("王五",23,32));
        arr.add(new Student("王二",30,44));
        Collections.sort(arr);
        System.out.println(arr);
    }

運行結果：
[Student{name=’王五’, age=23, secore=32}
, Student{name=’王二’, age=30, secore=44}
, Student{name=’張三’, age=18, secore=56}
, Student{name=’李四’, age=15, secore=57}
]
“`