要求
地铁系统跟踪不同车站之间的乘客出行时间，并使用这一数据来计算从一站到另一站的平均时间。
实现 UndergroundSystem 类：
- void checkIn(int id, string stationName, int t)
- 通行卡 ID 等于 id 的乘客，在时间 t ，从 stationName 站进入
- 乘客一次只能从一个站进入
- void checkOut(int id, string stationName, int t)
- 通行卡 ID 等于 id 的乘客，在时间 t ，从 stationName 站离开
- double getAverageTime(string startStation, string endStation)
- 返回从 startStation 站到 endStation 站的平均时间
- 平均时间会根据截至目前所有从 startStation 站 直接 到达 endStation 站的行程进行计算，也就是从 startStation 站进入并从 endStation 离开的行程
- 从 startStation 到 endStation 的行程时间与从 endStation 到 startStation 的行程时间可能不同
- 在调用 getAverageTime 之前，至少有一名乘客从 startStation 站到达 endStation 站 你可以假设对 checkIn 和 checkOut 方法的所有调用都是符合逻辑的。如果一名乘客在时间 t1 进站、时间 t2 出站，那么 t1 < t2 。所有时间都按时间顺序发生。
示例 1：
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,14.00000,11.00000,null,11.00000,null,12.00000]
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); // 乘客 45 "Leyton" -> "Waterloo" ，用时 15-3 = 12
undergroundSystem.checkOut(27, "Waterloo", 20); // 乘客 27 "Leyton" -> "Waterloo" ，用时 20-10 = 10
undergroundSystem.checkOut(32, "Cambridge", 22); // 乘客 32 "Paradise" -> "Cambridge" ，用时 22-8 = 14
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // 返回 14.00000 。只有一个 "Paradise" -> "Cambridge" 的行程，(14) / 1 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // 返回 11.00000 。有两个 "Leyton" -> "Waterloo" 的行程，(10 + 12) / 2 = 11
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // 返回 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38); // 乘客 10 "Leyton" -> "Waterloo" ，用时 38-24 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo"); // 返回 12.00000 。有三个 "Leyton" -> "Waterloo" 的行程，(10 + 12 + 14) / 3 = 12
示例 2：
输入
["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"]]
输出
[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]
解释
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(10, "Leyton", 3);
undergroundSystem.checkOut(10, "Paradise", 8); // 乘客 10 "Leyton" -> "Paradise" ，用时 8-3 = 5
undergroundSystem.getAverageTime("Leyton", "Paradise"); // 返回 5.00000 ，(5) / 1 = 5
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16); // 乘客 5 "Leyton" -> "Paradise" ，用时 16-10 = 6
undergroundSystem.getAverageTime("Leyton", "Paradise"); // 返回 5.50000 ，(5 + 6) / 2 = 5.5
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30); // 乘客 2 "Leyton" -> "Paradise" ，用时 30-21 = 9
undergroundSystem.getAverageTime("Leyton", "Paradise"); // 返回 6.66667 ，(5 + 6 + 9) / 3 = 6.66667
提示：
- 1 <= id, t <= 106
- 1 <= stationName.length, startStation.length, endStation.length <= 10 次
- 所有字符串由大小写英文字母与数字组成
- 总共最多调用 checkIn、checkOut 和 getAverageTime 方法 2 * 104
- 与标准答案误差在 10-5 以内的结果都被视为正确结果
核心代码
class UndergroundSystem:
def __init__(self):
from collections import defaultdict
self.cnt = defaultdict(int)
self.total = defaultdict(int)
self.record = {}
def checkIn(self, id: int, stationName: str, t: int) -> None:
self.record[id] = (stationName,t)
def checkOut(self, id: int, stationName: str, t: int) -> None:
checkinStation = self.record[id][0]
checkinTime = self.record[id][1]
self.total[checkinStation + "#" + stationName] += t - checkinTime
self.cnt[checkinStation + "#" + stationName] += 1
def getAverageTime(self, startStation: str, endStation: str) -> float:
return self.total[startStation + "#" + endStation] * 1.0 / self.cnt[startStation + "#" + endStation]
# 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)
解题思路：还是根据实际的题意设计相对应的数据结构实现相对应的功能，只有在getAverageTime的时候，需要两个数据，一个是起点到终点的总时间，一个是起点到终点的总人数，可得平均时间，就是dict的运用，比较简单。
