是否有可能/正确(to)或使用lambda重写下面的内容？在这里,我为KeyMapper和ValueMapper函数提供了内联实现.

public Map> getSubordinateHighestSalEmpMapV1(List employees) {

return employees.stream()

.filter(e -> e.getSubordinates() != null)

.collect(Collectors.toMap( //keyMapper

new Function() {

@Override

public Integer apply(Employee t) {

return t.getId();

}

},

new Function>() {//valueMapper

@Override

public List apply(Employee t) {

List subordinates = t.getSubordinates();

List subOrdinatesListWithHighestSalary = new ArrayList<>();

int maxSal = Integer.MIN_VALUE;

for(Employee s: subordinates) {

if(s.getSalary() >= maxSal) {

maxSal = s.getSalary();

}

}

for(Employee s: subordinates) {

if(s.getSalary() == maxSal) {

subOrdinatesListWithHighestSalary.add(s);

}

}

return subOrdinatesListWithHighestSalary;

}

}));

}

我想要实现的目标是什么：

员工类具有List< Employee>下属.我想在每个员工的下属中获得最高薪水.每个员工可能有也可能没有下属.如果下属不在,则不包括在结果中.如果不止一个下属具有相同的最高薪水,则所有下属都应该出现在结果中.

例如,它类似于获得每个部门中收入最高的员工(员工,如果工资匹配).

Employee.java

import java.util.List;

public class Employee{

private int id;

private int salary;

private List subordinates;

private String name;

private int age;

public int getId() {

return id;

}

public Employee setId(int id) {

this.id = id;

return this;

}

public int getSalary() {

return salary;

}

public Employee setSalary(int salary) {

this.salary = salary;

return this;

}

public List getSubordinates() {

return subordinates;

}

public Employee setSubordinates(List subordinates) {

this.subordinates = subordinates;

return this;

}

public String getName() {

return name;

}

public Employee setName(String name) {

this.name = name;

return this;

}

public int getAge() {

return age;

}

public Employee setAge(int age) {

this.age = age;

return this;

}

@Override

public String toString() {

return "Employee [id=" + id + ", salary=" + salary + ", name=" + name

+ ", age=" + age + "]";

}

@Override

public int hashCode() {

final int prime = 31;

int result = 1;

result = prime * result + id;

return result;

}

@Override

public boolean equals(Object obj) {

if (this == obj)

return true;

if (obj == null)

return false;

if (getClass() != obj.getClass())

return false;

Employee other = (Employee) obj;

if (id != other.id)

return false;

return true;

}

}

例如,对于以下输入：

> employee1(id：100)有employee2,employee3,employee4等等

employee3支付最高30000,应该是产出的一部分

> employee2(id：101)有employee5,employee6,其中,employee5的最高薪酬是20000,应该是输出的一部分

> employee3(id：102)有employee7和employee8,两者的薪水都是16000,输出应该包含两者.

> employee8(id：107)有一个下属employee9,薪水12000,employee9应该是输出的一部分

如下所述输入如下：

private static List getEmployeeListV1() {

int i = 100;

Employee employee1 = (Employee) new Employee().setId(i++).setSalary(10000).setAge(101).setName("emp 1");

Employee employee2 = (Employee) new Employee().setId(i++).setSalary(20000).setAge(110).setName("emp 2");

Employee employee3 = (Employee) new Employee().setId(i++).setSalary(30000).setAge(20).setName("emp 3");

Employee employee4 = (Employee) new Employee().setId(i++).setSalary(10000).setAge(32).setName("emp 4");

Employee employee5 = (Employee) new Employee().setId(i++).setSalary(20000).setAge(34).setName("emp 5");

Employee employee6 = (Employee) new Employee().setId(i++).setSalary(15000).setAge(44).setName("emp 6");

Employee employee7 = (Employee) new Employee().setId(i++).setSalary(16000).setAge(56).setName("emp 7");

Employee employee8 = (Employee) new Employee().setId(i++).setSalary(16000).setAge(65).setName("emp 8");

Employee employee9 = (Employee) new Employee().setId(i++).setSalary(12000).setAge(74).setName("emp 9");

employee1.setSubordinates(Stream.of(employee2,employee3,employee4).collect(Collectors.toList()));

employee2.setSubordinates(Stream.of(employee5,employee6).collect(Collectors.toList()));

employee3.setSubordinates(Stream.of(employee7,employee8).collect(Collectors.toList()));

employee8.setSubordinates(Stream.of(employee9).collect(Collectors.toList()));

List employees = Stream.of(employee1,employee2,

employee3,employee4,employee5,

employee6,employee7,employee8,

employee9).collect(Collectors.toList());

return employees;

}

以下是输出：

100=[Employee [id=102, salary=30000, name=emp 3, age=20]]

101=[Employee [id=104, salary=20000, name=emp 5, age=34]]

102=[Employee [id=106, salary=16000, name=emp 7, age=56], Employee [id=107, salary=16000, name=emp 8, age=65]]

107=[Employee [id=108, salary=12000, name=emp 9, age=74]]

阐释：