LeetCode 1396. Design Underground System Design subway system [Medium] [Python] [design]
Problem
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 stationstationNameat 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 stationstationNameat timet.
3. getAverageTime(string startStation, string endStation)
- Returns the average time to travel between the
startStationand theendStation. - The average time is computed from all the previous traveling from
startStationtoendStationthat happened directly. - Call to
getAverageTimeis always valid.
You can assume all calls to checkIn and checkOut 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,14.0,11.0,null,11.0,null,12.0]
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 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.0. 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 = 11.0
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.0
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.0
Constraints:
- There will be at most
20000operations. 1 <= id, t <= 10^6- All strings consist of uppercase, lowercase English letters and digits.
1 <= stationName.length <= 10- Answers within
10^-5of the actual value will be accepted as correct.
problem
You implement a class UndergroundSystem, it supports the following three methods:
1. checkIn(int id, string stationName, int t)
- Id number of passengers entering the metro station stationName at time t.
- A passenger can enter or leave a subway station at the same time.
2. checkOut(int id, string stationName, int t)
- Id number of passengers to leave the subway station stationName at time t.
3.getAverageTime(string startStation, string endStation)
- StartStation returned from the subway station to subway station endStation spend an average time of.
- The itinerary includes the calculated average time until the current reaches all endStation travel directly from startStation.
- When you call getAverageTime, ask the route contains at least a trip itinerary.
You may assume that all calls to checkIn and checkOut are logical. That is, if a customer arrives at a subway station at the time t1, then he left to meet certain time t2 t2> t1. All events are given in chronological order.
Example:
输入：
["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"]]
输出：
[null,null,null,null,null,null,null,14.0,11.0,null,11.0,null,12.0]
解释：
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"); // 返回 14.0。从 "Paradise"（时刻 8）到 "Cambridge"(时刻 22)的行程只有一趟
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // 返回 11.0。总共有 2 躺从 "Leyton" 到 "Waterloo" 的行程，编号为 id=45 的乘客出发于 time=3 到达于 time=15，编号为 id=27 的乘客于 time=10 出发于 time=20 到达。所以平均时间为 ( (15-3) + (20-10) ) / 2 = 11.0
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // 返回 11.0
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // 返回 12.0
prompt:
- A total of up to 20,000 operations.
- 1 <= id, t <= 10^6
- All string contains uppercase letters, lowercase letters and numbers.
- 1 <= stationName.length <= 10
- The answer is within the standard error of 10 ^ -5 the results are deemed to be correct results.
Thinking
Design issues
字典存储到站离站乘客
字典的 key 是站台名，value 是二维列表 [id, t]
Python3 Code
class UndergroundSystem:
def __init__(self):
self.enterstation = {}
self.leavestation = {}
def checkIn(self, id: int, stationName: str, t: int) -> None:
if stationName not in self.enterstation:
self.enterstation[stationName] = [[id, t]]
else:
self.enterstation[stationName].append([id, t])
def checkOut(self, id: int, stationName: str, t: int) -> None:
if stationName not in self.leavestation:
self.leavestation[stationName] = [[id, t]]
else:
self.leavestation[stationName].append([id, t])
def getAverageTime(self, startStation: str, endStation: str) -> float:
res = []
start = self.enterstation[startStation]
end = self.leavestation[endStation]
for i in start:
for j in end:
# id相同
if i[0] == j[0]:
res.append(abs(j[1] - i[1]))
return float(sum(res) / len(res))
# 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)