使用兩個列表進行流查詢

Question

我有一個Visit對象列表，現在我想構建另一個列表，其中包含給定日期的可用小時數;

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

這是包含所有訪問時間的數組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"};

所以現在我從Db訪問列表中獲取（每次訪問都定義了時間），並檢查是否還有其他空閑時間來安排訪問。

我已經編寫了兩種方法，其中一種方法是迭代，第二種是使用流，兩種方法都按預期工作。

我問的是如何重建這個方法不使用終端方法兩次。

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());
    }

這里是收藏的老派方法：

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

你可以試試這個。 你在這里獲得了一些好處，與List相比， HashSet contains的速度更快

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;
}