LeetCode 1396. Design Underground System设计地铁系统【Medium】【Python】【设计】
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 12.0Constraints:
- 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.
问题
请你实现一个类 UndergroundSystem ，它支持以下 3 种方法：
1.checkIn(int id, string stationName, int t)
- 编号为 id 的乘客在 t 时刻进入地铁站 stationName 。
- 一个乘客在同一时间只能在一个地铁站进入或者离开。
2.checkOut(int id, string stationName, int t)
- 编号为 id 的乘客在 t 时刻离开地铁站 stationName 。
3.getAverageTime(string startStation, string endStation)
- 返回从地铁站 startStation 到地铁站 endStation 的平均花费时间。
- 平均时间计算的行程包括当前为止所有从 startStation 直接到达 endStation 的行程。
- 调用 getAverageTime 时，询问的路线至少包含一趟行程。
你可以假设所有对 checkIn 和 checkOut 的调用都是符合逻辑的。也就是说，如果一个顾客在 t1 时刻到达某个地铁站，那么他离开的时间 t2 一定满足 t2 > t1 。所有的事件都按时间顺序给出。
示例：
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提示：
- 总共最多有 20000 次操作。
- 1 <= id, t <= 10^6
- 所有的字符串包含大写字母，小写字母和数字。
- 1 <= stationName.length <= 10
- 与标准答案误差在 10^-5 以内的结果都视为正确结果。
思路
设计题
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代码
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) 
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