题目
错误解法
先分享一种错误解法。下面的这种错误解法错在很多个点上。首先,他每个操作不是O(1)的,时间复杂度就错了。第二,如果某个id先进出某两个站,然后再进了某个站,这个时候计算他的travel time是错的,因为按照下面这种解法会是出站时间小于进站时间,也就是这种解法不满足同一个id多次进出的情况。第三,如果某个idA进B出,然后C进D出,这个时候按下面的解法,在query(A,D)这个pair的时候就会被错误的包括进去,所以start和end必须是以pair的形式储存的
class UndergroundSystem:
def __init__(self):
self.checkin_dict = {}
self.checkout_dict = {}
def checkIn(self, id: int, stationName: str, t: int) -> None:
if stationName in self.checkin_dict:
self.checkin_dict[stationName][id] = t
else:
self.checkin_dict[stationName] = {}
self.checkin_dict[stationName][id] = t
def checkOut(self, id: int, stationName: str, t: int) -> None:
if stationName in self.checkout_dict:
self.checkout_dict[stationName][id] = t
else:
self.checkout_dict[stationName] = {}
self.checkout_dict[stationName][id] = t
def getAverageTime(self, startStation: str, endStation: str) -> float:
#print(self.checkin_dict)
total_time = 0
cnt = 0
if startStation in self.checkin_dict and endStation in self.checkout_dict:
for id,t in self.checkin_dict[startStation].items():
if id in self.checkout_dict[endStation]:
cnt += 1
total_time += self.checkout_dict[endStation][id] - self.checkin_dict[startStation][id]
return total_time/cnt
正确解法
这道题正确解法远没有这么复杂。前面这么些是因为默认了checkin和checkout的情况需要被储存下来,其实只要储存checkin,checkout的时候就可以直接计算相应的时间了
class UndergroundSystem:
def __init__(self):
self.checkin_dict = {}
self.travel_time = collections.defaultdict(int)
self.travel_cnt = collections.defaultdict(int)
def checkIn(self, id: int, stationName: str, t: int) -> None:
self.checkin_dict[id] = (stationName,t)
def checkOut(self, id: int, stationName: str, t: int) -> None:
startStation,start_t = self.checkin_dict[id]
self.travel_time[(startStation,stationName)] += t - start_t
self.travel_cnt[(startStation,stationName)] += 1
def getAverageTime(self, startStation: str, endStation: str) -> float:
return self.travel_time[(startStation,endStation)]/self.travel_cnt[(startStation,endStation)]