如何在列表<对象>中找到最大日期？

collections java

2022-08-31 12:56:58

考虑一个类User

public class User{
  int userId;
  String name;
  Date date;
}

现在我有一个大小为20的大小，如何在不使用手动迭代器的情况下找到列表中的最大日期？List<User>

答案 1

由于您要求 lambdas，因此您可以在 Java 8 中使用以下语法：

Date maxDate = list.stream().map(u -> u.date).max(Date::compareTo).get();

或者，如果您有日期的获取者：

Date maxDate = list.stream().map(User::getDate).max(Date::compareTo).get();

答案 2

与接受的答案相比，一个小小的改进是执行空检查并获得完整的对象。

public class DateComparator {

public static void main(String[] args) throws ParseException {
    SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd");

    List<Employee> employees = new ArrayList<>();
    employees.add(new Employee(1, "name1", addDays(new Date(), 1)));
    employees.add(new Employee(2, "name2", addDays(new Date(), 3)));
    employees.add(new Employee(3, "name3", addDays(new Date(), 6)));
    employees.add(new Employee(4, "name4", null));
    employees.add(new Employee(5, "name5", addDays(new Date(), 4)));
    employees.add(new Employee(6, "name6", addDays(new Date(), 5)));
    System.out.println(employees);
    Date maxDate = employees.stream().filter(emp -> emp.getJoiningDate() != null).map(Employee::getJoiningDate).max(Date::compareTo).get();
    System.out.println(format.format(maxDate));
    //Comparator<Employee> comparator = (p1, p2) -> p1.getJoiningDate().compareTo(p2.getJoiningDate());
    Comparator<Employee> comparator = Comparator.comparing(Employee::getJoiningDate);
    Employee maxDatedEmploye = employees.stream().filter(emp -> emp.getJoiningDate() != null).max(comparator).get();
    System.out.println(" maxDatedEmploye : " + maxDatedEmploye);

    Employee minDatedEmployee = employees.stream().filter(emp -> emp.getJoiningDate() != null).min(comparator).get();
    System.out.println(" minDatedEmployee : " + minDatedEmployee);

}

public static Date addDays(Date date, int days) {
    Calendar cal = Calendar.getInstance();
    cal.setTime(date);
    cal.add(Calendar.DATE, days); // minus number would decrement the days
    return cal.getTime();
}
}

你会得到以下结果：

 [Employee [empId=1, empName=name1, joiningDate=Wed Mar 21 13:33:09 EDT 2018],
Employee [empId=2, empName=name2, joiningDate=Fri Mar 23 13:33:09 EDT 2018],
Employee [empId=3, empName=name3, joiningDate=Mon Mar 26 13:33:09 EDT 2018],
Employee [empId=4, empName=name4, joiningDate=null],
Employee [empId=5, empName=name5, joiningDate=Sat Mar 24 13:33:09 EDT 2018],
Employee [empId=6, empName=name6, joiningDate=Sun Mar 25 13:33:09 EDT 2018]
]
2018-03-26
 maxDatedEmploye : Employee [empId=3, empName=name3, joiningDate=Mon Mar 26 13:33:09 EDT 2018]

 minDatedEmployee : Employee [empId=1, empName=name1, joiningDate=Wed Mar 21 13:33:09 EDT 2018]

更新：如果列表本身是空的怎么办？

        Date maxDate = employees.stream().filter(emp -> emp.getJoiningDate() != null).map(Employee::getJoiningDate).max(Date::compareTo).orElse(new Date());
    System.out.println(format.format(maxDate));
    Comparator<Employee> comparator = Comparator.comparing(Employee::getJoiningDate);
    Employee maxDatedEmploye = employees.stream().filter(emp -> emp.getJoiningDate() != null).max(comparator).orElse(null);
    System.out.println(" maxDatedEmploye : " + maxDatedEmploye);