这里的主要任务是检索名称长度大于3的雇员的ID。但是雇员4的名称为null，因此将引发null指针异常。如何跳过引发异常并处理列表中剩余元素而不是终止列表的employee 4。期望的输出可以是[1、2、3、5、6、7、8]
代码如下：

public class EmployeeTest {
    public static void main(String[] args) {
        List<Employee> empList = new ArrayList<>();
        createEmpList(empList);
        List<Integer> employeeIds = empList.stream()
                .filter(x -> x.getName().length() > 3)
                .map(x -> x.getId())
                .collect(Collectors.toList());
        System.out.println(employeeIds);
    }

    private static void createEmpList(List<Employee> empList) {
        Employee e1 = new Employee("siddu",   1, "Hyderabad", 70000);
        Employee e2 = new Employee("Swami",   2, "Hyderabad", 50000);
        Employee e3 = new Employee("Ramu",    3, "Bangalore", 100000);
        Employee e4 = new Employee(null,      4, "Hyderabad", 65000);
        Employee e5 = new Employee("Krishna", 5, "Bangalore", 160000);
        Employee e6 = new Employee("Naidu",   6, "Poland",    250000);
        Employee e7 = new Employee("Arun",    7, "Pune",      45000);
        Employee e8 = new Employee("Mahesh",  8, "Chennai",   85000);

        empList.add(e1);
        empList.add(e2);
        empList.add(e3);
        empList.add(e4);
        empList.add(e5);
        empList.add(e6);
        empList.add(e7);
        empList.add(e8);
    }
}