如何在List 中找到Max Date？

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List ，如何在不使用手动迭代器的情况下在列表中找到最大日期 ？

Answer 1

由于您要求使用lambdas，因此可以在Java 8中使用以下语法：

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

或者，如果你有一个吸气剂的日期：

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

Answer 2

 Comparator cmp = new Comparator() { @Override public int compare(User user1, User user2) { return user1.date.compareTo(user2.date); } }; Collections.max(list, cmp); 

Answer 3

接受答案的一个小改进是进行空检查并获得完整对象。

 public class DateComparator { public static void main(String[] args) throws ParseException { SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd"); List 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 comparator = (p1, p2) -> p1.getJoiningDate().compareTo(p2.getJoiningDate()); Comparator 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：如果列表本身为空怎么办？

  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 comparator = Comparator.comparing(Employee::getJoiningDate); Employee maxDatedEmploye = employees.stream().filter(emp -> emp.getJoiningDate() != null).max(comparator).orElse(null); System.out.println(" maxDatedEmploye : " + maxDatedEmploye);