【leetcode】1396. 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 station stationName at time t.
• 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 station stationName at time t.

3. getAverageTime(string startStation, string endStation) 

• Returns the average time to travel between the startStation and the endStation.
• The average time is computed from all the previous traveling from startStation to endStation that happened directly.
• Call to getAverageTime is 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,14.00000,11.00000,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 12.00000

Example 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 6.66667

Constraints:

• 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)

【leetcode】1396. Design Underground System

标签：type   value   within   accept   length   not   des   sum   mbr   

原文地址：https://www.cnblogs.com/seyjs/p/12631557.html