compareTo 方法逻辑按多个变量对 List 进行排序

Question

问题：创建一个独立的 jar 可执行文件，它将按姓名、年龄和经验升序打印出现在面试中的候选人列表。

我无法弄清楚 compareTo 方法逻辑以便能够对给定问题中的 3 个字段进行排序。

员工类

package com.example.demo.employee;

public class Employee implements Comparable<Employee> {

private String name;
private int age;
private int exp;

public Employee(String name, int age, int exp) {
    super();
    this.name = name;
    this.age = age;
    this.exp = exp;
}

public Employee() {
}

// getter setter

@Override
public int compareTo(Employee emp) {

    // I do not think this logic is correct
    // I have read the other stack overflow posts with similar problem
    // but failing to under stand what to do in this method.

    int result = (this.name).compareTo(emp.name);
    if ( result == 0 ) {
        result = (this.age).compareTo(emp.age);
    }

    if ( result == 0 ) {
        result = (this.exp).compareTo(emp.exp);
    }
    return result;
 }

}

员工服务类

package com.example.demo.employee;

import org.springframework.stereotype.Service;

import java.util.List;

@Service
public class EmployeeService {


public List<Employee> getEmployees() {

    Employee e1 = new Employee("Sandhya", 20, 0);
    Employee e2 = new Employee("Kemp", 24, 2);
    Employee e3 = new Employee("Anil", 22, 3);
    Employee e4 = new Employee("Kumar", 30, 6);
    Employee e5 = new Employee("Tim", 32, 7);

public List<Employee> getEmployees() {

    List<Employee> eList = new ArrayList<>();
    eList.add(e1);
    eList.add(e2);
    eList.add(e3);
    eList.add(e4);
    eList.add(e5);

    Collections.sort(eList);

    return eList;
  }
}

员工控制器

package com.example.demo.employee;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RestController;

import java.util.List;

@RestController
public class EmployeeController {

@Autowired
EmployeeService es;

@RequestMapping(value = "/")
public List<Employee> getEmpList(){
    List<Employee> list = es.getEmployees();
    return list;
  }

}

Answer 1

public class Employee implements Comparable<Employee> {

    private String name;
    private int age;
    private int exp;
    public int getAge() {
        return age;
    }
    public void setAge(int age) {
        this.age = age;
    }
    public int getExp() {
        return exp;
    }
    public void setExp(int exp) {
        this.exp = exp;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Employee(String name, int age, int exp) {
        super();
        this.name = name;
        this.age = age;
        this.exp = exp;
    }
    public Employee() {
  }

    @Override
    public int compareTo(Employee emp) {
    //replace your comparator here
       return (this.getAge() - emp.getAge());
  }

}
-------------------------------------------------------
@Service
public class EmployeeService {


    public List<Employee> getEmployees() {
        List<Employee> emp = new ArrayList<>();
        Employee emp1 = new Employee("Sandhya",20,0);
        Employee emp2 = new Employee("Kemp",24,2);
        Employee emp3 = new Employee("Anil",22,3);
        Employee emp4 = new Employee("Kumar",30,6);
        Employee emp5 = new Employee("Tim",32,7);
        emp.add(emp1);
        emp.add(emp2);
        emp.add(emp3);
        emp.add(emp4);
        emp.add(emp5);

        return emp;
    }




}
---------------------------------------------------------------------------

@RestController
public class EmployeeController {

  @Autowired
  EmployeeService employeeService;

  @RequestMapping("/")
    public List<Employee> getEmpList(){
    List<Employee> empController = employeeService.getEmployees();
    Collections.sort(empController);
        return empController;
    }

}

Answer 2

无需实现Comparable并重写compareTo方法，使用Comparator即可

Comparator<Employee> c = Comparator.comparing(Employee::getName)
                                       .thenComparing(Employee::getAge)
                                       .thenComparing(Employee::getExp);

并且使用Collections.sort() 使用传递的Comparator 对列表进行排序

Collections.sort(eList,c);

通过使用 Comparable

问题是age 和exp 是int 类型，这是原始类型，如果您不能使用compareTo 方法，请将它们的类型更改为Integer 包装对象或使用Integer.compare(int a, int b) 方法

private int age;    // to private Integer age
private int exp;    // to private Integer exp

这样您就可以在age 和exp 上使用compareTo

this.getAge().compareTo(o.getAge());
this.getExp().compareTo(o.getExp());

如果不使用Integer.compare(int a, int b)查看我的解决方案

解决方案

@Override
public int compareTo(Employee o) {
    int result = this.getName().compareTo(o.getName());
    if (result == 0) {
        result = Integer.compare(this.getAge(), o.getAge());
        if (result == 0) {
            return Integer.compare(this.getExp(), o.getExp());
        }
        return result;
    }
    return result;
}

Answer 3

因此，问题似乎是您的 compareTo 中没有发生自动装箱，因此您可以将它们装箱，或者仅将 age/exp 作为原始整数处理。 所以这样做，将原始 int 转换为 Integer 具有 compareTo:

public int compareTo(Employee emp) {
    int result = (this.name).compareTo(emp.name);
    if ( result == 0 ) {
        result = Integer.valueOf(age).compareTo(emp.age);
    }
    if ( result == 0 ) {
        result = Integer.valueOf(exp).compareTo(emp.exp);
    }
    return result;
 }

或者，您也可以这样做，只保留原语：

public int compareTo(Employee emp) {
    int result = (this.name).compareTo(emp.name);
    if ( result == 0 ) {
        result = Integer.compare( age, emp.age );
    }
    if ( result == 0 ) {
        result = Integer.compare( exp, emp.exp );
    }
    return result;
 }

Answer 4

您的代码看起来非常好，我不得不思考一下为什么它可能是错误的。

关键是int 和String 在Java 中的类型非常不同。 int 是所谓的原始类型（因为它不是由其他类型组成的），String 是对象类型。按照惯例，原始类型的名称以小写字母开头，而对象类型的名称以大写字母开头。

只有对象类型可以有方法。

由于this.age 的类型是int，因此是原始类型，因此不允许使用(this.age).compareTo(...)。相反，你必须写Integer.compare(this.age, emp.age)。

编译器的错误信息并没有真正的帮助。与其说“方法 int.compareTo 未找到”，不如说“this.age 的类型是 int，因为它是原始类型，所以不能对其调用任何方法”。

Answer 5

对你的排序问题代码的正确修改是

//In service class

public List<Employee> getEmployees() {

        Employee e1 = new Employee("Sandhya", 24, 0);
        Employee e2 = new Employee("Kemp", 22, 2);
        Employee e3 = new Employee("Anil", 26, 3);
        Employee e4 = new Employee("Kumar", 30, 6);
        Employee e5 = new Employee("Tim", 32, 7);

        List<Employee> elist = Arrays.asList(e1,e2,e3,e4,e5);

        return elist;

//in controller class

@Autowired
    EmployeeService empserv;
    
    @RequestMapping(value="/")
    public List<Employee> getEmpList(){
        List<Employee> list = empserv.getEmployees();
        Collections.sort(list);
        return list;
    }


//in Employee class, since it was asked to sort according to age of the employees

@Override

    public int compareTo(Employee o) {
        int result = Integer.compare(this.getAge(), o.getAge());;
        /*int result = this.getName().compareTo(o.getName());
        if (result == 0) {
            result = Integer.compare(this.getAge(), o.getAge());
            if (result == 0) {
                return Integer.compare(this.getExp(), o.getExp());
            }
            return result;
        }*/
    
        return result;
    }

Answer 6

通过这个测试。

在Employee类中添加

public int compareTo(Employee emp) { 返回 this.getAge() - emp.getAge();

}

在 EmployeeService 中，只需在 getEmployess 方法中从 TestEmployee 类中复制列表数据并返回列表。

在EmployeeController 中添加@RequestMapping("/") 并使用Collections.sort(list) 并返回列表；