leetcode1396. Design Underground System
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题目如下:
Implement the class
UndergroundSystem
that supports three methods:1.
checkIn(int id, string stationName, int t)
- A customer with id card equal to
id
, gets in the stationstationName
at timet
.- A customer can only be checked into one place at a time.
2.
checkOut(int id, string stationName, int t)
- A customer with id card equal to
id
, gets out from the stationstationName
at timet
.3.
getAverageTime(string startStation, string endStation)
- Returns the average time to travel between the
startStation
and theendStation
.- The average time is computed from all the previous traveling from
startStation
toendStation
that happened directly.- Call to
getAverageTime
is always valid.You can assume all calls to
checkIn
andcheckOut
methods are consistent. That is, if a customer gets in at time t1 at some station, then it gets out at time t2 with t2 > t1. All events happen in chronological order.Example 1:
Input
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"] [[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]] Output [null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000] Explanation UndergroundSystem undergroundSystem = new UndergroundSystem(); undergroundSystem.checkIn(45, "Leyton", 3); undergroundSystem.checkIn(32, "Paradise", 8); undergroundSystem.checkIn(27, "Leyton", 10); undergroundSystem.checkOut(45, "Waterloo", 15); undergroundSystem.checkOut(27, "Waterloo", 20); undergroundSystem.checkOut(32, "Cambridge", 22); undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22) undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.00000 undergroundSystem.checkIn(10, "Leyton", 24); undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000 undergroundSystem.checkOut(10, "Waterloo", 38); undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.00000Example 2:
Input
["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"] [[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]] Output [null,null,null,5.00000,null,null,5.50000,null,null,6.66667] Explanation UndergroundSystem undergroundSystem = new UndergroundSystem(); undergroundSystem.checkIn(10, "Leyton", 3); undergroundSystem.checkOut(10, "Paradise", 8); undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000 undergroundSystem.checkIn(5, "Leyton", 10); undergroundSystem.checkOut(5, "Paradise", 16); undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000 undergroundSystem.checkIn(2, "Leyton", 21); undergroundSystem.checkOut(2, "Paradise", 30); undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667Constraints:
- There will be at most
20000
operations.1 <= id, t <= 10^6
- All strings consist of uppercase, lowercase English letters and digits.
1 <= stationName.length <= 10
- Answers within
10^-5
of the actual value will be accepted as correct.
解题思路:在checkin的时候,用字典记录每一个乘客的上车车站,key值为乘客id;checkout的时候,根据乘客id找出上车车站,同时用另一个字典记录乘客的乘车时间,key值为(上车车站,下车车站),同时记录key值出现的次数。
代码如下:
class UndergroundSystem(object): def __init__(self): self.dic_passagner = {} self.dic_station = {} def checkIn(self, id, stationName, t): """ :type id: int :type stationName: str :type t: int :rtype: None """ self.dic_passagner[id] = [stationName,t] def checkOut(self, id, stationName, t): """ :type id: int :type stationName: str :type t: int :rtype: None """ checkIn_station,checkIn_time = self.dic_passagner[id] del self.dic_passagner[id] if (checkIn_station,stationName) not in self.dic_station: self.dic_station[(checkIn_station,stationName)] = [t-checkIn_time,1] else: self.dic_station[(checkIn_station, stationName)][0] += t - checkIn_time self.dic_station[(checkIn_station, stationName)][1] += 1 def getAverageTime(self, startStation, endStation): """ :type startStation: str :type endStation: str :rtype: float """ return float(self.dic_station[(startStation,endStation)][0]) / float(self.dic_station[(startStation,endStation)][1]) # Your UndergroundSystem object will be instantiated and called as such: # obj = UndergroundSystem() # obj.checkIn(id,stationName,t) # obj.checkOut(id,stationName,t) # param_3 = obj.getAverageTime(startStation,endStation)
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