将循环转换为 Java 流

Question

我正在编写这个逻辑，我正在为每个员工分配单独的房间，我将其代码粘贴在下面：

public List<EmployeeRooms> assignRoomEmployee(Request request) {
    List<Employee> allEmployeeList = getAllEmployeeList(request);
    int possibleAssigments = Math.min(request.getRooms().size(), allEmployeeList.size());        
    List<EmployeeRooms> finalList = new ArrayList<>();
    int i = 0;
    while(i < possibleAssigments) {                                         
     for (int j= 0; j < allEmployeeList.size(); j++) {                       
         finalList.add(createRoomEmployeeList(allEmployeeList.get(j), request.getRooms().get(i)));                                       
         i++;
         }          
    }
    return finalList;
    
}

现在我想使用单个 Java 流语句编写此循环逻辑，但我无法正确执行此操作，我尝试了下面的代码，但它将每个房间分配给每个员工，从而创建重复而不是每个房间给单个员工：

 while(i < possibleAssigments) {             
     allEmployeeList.stream()
                    .map(emp -> createRoomEmployeeList(emp, request.getRooms().get(i)))
                    .collect(Collectors.toList());
     i++;
 }

Answer 1

您可以使用IntStream()在i和possibleAssigments之间循环，然后为每次迭代返回一个RoomEmployeeList ：

x -> createRoomEmployeeList(allEmployeeList.get(x), request.getRooms().get(x)))

最后收集它们。

我没有测试，但它应该是这样的：

finalList = IntStream.range(i, possibleAssigments).boxed().map(
                x -> createRoomEmployeeList(allEmployeeList.get(x), request.getRooms().get(x)))
                .collect(Collectors.toList());

assignRoomEmployee()方法：

public List<EmployeeRooms> assignRoomEmployee(Request request) {
    List<Employee> allEmployeeList = getAllEmployeeList(request);
    int possibleAssigments = Math.min(request.getRooms().size(), allEmployeeList.size());
    int i = 0;
    
    List<EmployeeRooms> finalList = IntStream.range(i, possibleAssigments).boxed().map(
            x -> createRoomEmployeeList(allEmployeeList.get(x), request.getRooms().get(x)))
            .collect(Collectors.toList());

    return finalList;
}

Answer 2

我看到的唯一错误是您没有将此 stream 操作的结果保存在变量中。

这应该是

 List<EmployeeRooms> finalList = new ArrayList<>();
 while(i < possibleAssigments) {             
     finalList.addAll(allEmployeeList.stream().map(emp -> createRoomEmployeeList(emp, request.getRooms().get(i))).collect(Collectors.toList()));
     i++;
 }

Answer 3

为员工分配房间

• 迭代 0 到possibleAssigments-1
• 获取员工allEmployeeList.get(i)和房间request.getRooms().get(i)
• 创建EmployeeRooms

使用 Stream API 您可以这样做。

List<EmployeeRooms> finalList = 
    IntStream.range(0, possibleAssigments)
             .boxed()
             .map(i -> createRoomEmployeeList(allEmployeeList.get(i),
                                              request.getRooms().get(i)))
             .collect(Collectors.toList());