LeetCode | 1396. Design Underground System设计地铁系统【Python】

LeetCode 1396. Design Underground System设计地铁系统【Medium】【Python】【设计】

Problem

LeetCode

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,14.0,11.0,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 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.

问题

力扣

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)

GitHub链接

Python

参考

python3 字典存储

posted @ 2020-03-29 14:03  Wonz  阅读(275)  评论(0编辑  收藏  举报