我有一个Visit对象列表,现在我想构建另一个列表,其中包含给定日期的可用小时数;
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"};
所以现在我从Db访问列表中获取(每次访问都定义了时间),并检查是否还有其他空闲时间来安排访问。
我已经编写了两种方法,其中一种方法是迭代,第二种是使用流,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;
}
答案 0 :(得分:1)
你可以试试这个。您可以在此获得一些好处,与{List
相比,contains
的{{1}}速度更快
HashSet