java - 使用两个列表进行流式查询

Question

我有一个 Visit 对象列表，现在我想构建另一个包含给定日期可用时间的列表；

public class Visit {
    private int id;
    private Date date;
    private Time time;
    private Pet pet;
    private Vet vet;  

这是包含所有访问时间的数组 String[]:

public class VisitTime {

    private static final String[] visitTime = 
        {"09:00:00","09:30:00","10:00:00","10:30:00","11:00:00","11:30:00","12:00:00",
                "12:30:00","13:00:00","13:30:00","14:00:00","14:30:00","15:00:00","15:30:00","16:00:00","16:30:00"};

现在我从 Db 获取访问列表(每次访问都有定义的时间)，并检查是否有任何其他空闲时间来安排访问。

我已经编写了两种方法来做到这一点，一种是使用流进行迭代，两种方法都按预期工作。

我要问的是如何重建此方法以不使用终端方法两次。

public List<String> getHoursAvailable12(int vetId, String date){
        List<Visit> visitList = getVisitByVetIdAndDate(vetId, date);
        List<String> hoursAvailable = new ArrayList<>(Arrays.asList(VisitTime.getVisittime()));


        List<String> hoursTaken = visitList.stream().map(Visit::getTime).map(Time::toString).collect(Collectors.toList());

        return hoursAvailable.stream().filter(x -> !hoursTaken.contains(x)).collect(Collectors.toList());
    }

这是老式的集合方法:

public List<String> getHoursAvailable(int vetId, String date){
        List<Visit> visitList = getVisitByVetIdAndDate(vetId,date);
        ArrayList<String> hoursAvailable = new ArrayList<>(Arrays.asList(VisitTime.getVisittime()));

        for(Visit x : visitList){
            {
            String time = x.getTime().toString();
            if(hoursAvailable.contains(time)) hoursAvailable.remove(time);
            }
        }
        return hoursAvailable;
    }

Answer 1

你可以试试这个。你在这里得到了一些好处，contains 在 HashSet 中比 List

更快

public Set<String> getHoursAvailable(int vetId, String date){
    List<Visit> visitList = getVisitByVetIdAndDate(vetId,date);
    Set<String> hoursAvailable = new LinkedHashSet<>(
          Arrays.asList(VisitTime.getVisittime()));

    visitList.stream()
       .map(Visit::getTime)
       .map(Time::toString)
       .forEach(vt-> hoursAvailable.removeIf(s->s.equals(vt)));

    return hoursAvailable;
}