我们假设我有一个Flight
对象列表
public class Flight {
private int passengers;
private int price
...
//getters and Setters
}
List<Flight> flightList = new ArrayList<Flight>();
现在我需要累积每位乘客的价格和价格,因为我必须能够在以后进行这两种信息。所以我会创建两个方法:
public int calculatePrice(List<Flight> flightList) {
int price = 0;
for (Flight flight : flightList) {
price = price + flight.getPrice();
}
return price;
}
public int calculatePricePerPassengers(List<Flight> flightList) {
int pricePerPassenger = 0;
for (Flight flight : flightList) {
pricePerPassenger = (pricePerPassenger) + (flight.getPrice() / flight.getPassgers());
}
return pricePerPassenger;
}
我有4-5种相同类型的方法。我不确定是否有太多冗余,因为我调用for循环4-5次,我可以轻松地在一个for循环中执行所有操作,但这会产生多个返回值的效果。这是适当的方式吗? (只是一个虚拟的例子)
答案 0 :(得分:5)
您可以使用map或者也可以扭曲要在类中返回的所有字段。在这里,我将使用Map进行示例。
public static Map<String, Integer> calculatePrice(List<Flight> flightList) {
Map<String, Integer> priceMap = new HashMap<>();
int price = 0;
int pricePerPassenger = 0;
for (Flight flight : flightList) {
price = price + flight.getPrice();
pricePerPassenger = (pricePerPassenger) + (flight.getPrice() / flight.getPassgers());
}
priceMap.put("price", price);
priceMap.put("pricePerPassenger", pricePerPassenger);
return priceMap;
}
修改强> 使用包装类(DTO)的示例。
public static FligtPriceDTO calculatePrice(List<Flight> flightList) {
FligtPriceDTO dto = new FligtPriceDTO();
int price = 0;
int pricePerPassenger = 0;
for (Flight flight : flightList) {
price = price + flight.getPrice();
pricePerPassenger = (pricePerPassenger) + (flight.getPrice() / flight.getPassgers());
}
dto.setPrice(price);
dto.setPricePerPassenger(pricePerPassenger);
return dto;
}
}
class FligtPriceDTO {
private int price;
private int pricePerPassenger;
public int getPrice() {
return price;
}
public void setPrice(int price) {
this.price = price;
}
public int getPricePerPassenger() {
return pricePerPassenger;
}
public void setPricePerPassenger(int pricePerPassenger) {
this.pricePerPassenger = pricePerPassenger;
}
}
答案 1 :(得分:1)
您可以将这两个值包装到一个新数组中:
public int[] calculatePrices(List<Flight> flightList) {
int totalPrice = 0;
int pricePerCustomer = 0;
for (Flight flight : flightList) {
totalPrice += flight.getPrice();
pricePerCustomer += (flight.getPrice() / flight.getPassgers());
}
return new int[] { totalPrice, pricePerCustomer };
}
答案 2 :(得分:0)
这一切都归结为您希望为用户提供的功能。因此,如果用户希望能够单独计算两种价格,那么我看不到任何其他方式。但是,如果您可以同时提供两种价格,那么您可以消除其中一种方法,只在您的航班列表中循环一次并返回两种价格的组合。例如:
public static Pair[] calculatePrices(List<Flight> flightList)
{
int pricePerFlight= 0;
int pricePerPassenger = 0;
for(Flight flight : flightList)
{
pricePerFlight = pricePerFlight + flight.getPrice();
pricePerPassenger = pricePerPassenger +(flight.getPrice()/flight.getPassgers());
}
return new Pair[] {new Pair<>("pricePerFlight", pricePerFlight), new Pair<>("pricePerPassenger", pricePerPassenger )};
}