Question

Set<Employee> employeeSet =

如何从Set中选择最老的Employee（它有一个属性'int age'）

Answer 1

您可以使用Stream API

Optional<Employee> employee = employeeSet.stream()
                                         .max(Comparator.comparing(e -> e.age));

employee.get()将返回Employee，除非该集合为空。

Answer 2

这可能不是性能的最佳解决方案，但您可以使用增强的for循环遍历每个元素：

Employee oldest = employeeSet.get(0); // Initialize with the first index
for (Employee emp : employeeSet) {
    if (emp.age > oldest.age) {
        oldest = emp;
    }
}

Answer 3

我认为Java 8中的Stream接口可以提供帮助

final Comparator<Employee> comp = Comparator.comparingInt(Employee::getAge);
Employee oldest = employeeList.stream()
            .max(comp)
            .get();

详细了解此https://stevewall123.wordpress.com/2014/08/31/java-8-streams-max-and-streams-min-example/

Answer 4

您可以遍历该集合。

int maxAge = 0; //  
for(Employee emp: employeeSet){
    if(emp.getAge() > maxAge)
      maxAge = employee.getAge()
}

maxAge变量将在forEach循环结束后具有最高年龄

Answer 5

您可以为查找最早的员工定义第一个年龄 int baseAge = 50; 现在您需要将所有员工年龄与此年龄进行比较

  Public Set findOldest()
    {
        int baseAge =50;
        Set < Employee > oldest = new HashSet< Employee >();
        Set<Employee> employee = // your set of employees
        for ( Employee emp : employee )
        {
            If(emp.empAge > baseAge)
                    oldest.add(emp);
        }
        return oldest;
    }

根据其属性从Set中选择一个元素

5 个答案: