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155 lines
6.7 KiB
Markdown
155 lines
6.7 KiB
Markdown
# [1396. Design Underground System](https://leetcode.com/problems/design-underground-system/)
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## 题目
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Implement the `UndergroundSystem` class:
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- `void checkIn(int id, string stationName, int t)`
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- A customer with a card id equal to `id`, gets in the station `stationName` at time `t`.
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- A customer can only be checked into one place at a time.
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- `void checkOut(int id, string stationName, int t)`
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- A customer with a card id equal to `id`, gets out from the station `stationName` at time `t`.
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- `double getAverageTime(string startStation, string endStation)`
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- Returns the average time to travel between the `startStation` and the `endStation`.
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- The average time is computed from all the previous traveling from `startStation` to `endStation` that happened **directly**.
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- Call to `getAverageTime` is always valid.
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You can assume all calls to `checkIn` and `checkOut` methods are consistent. If a customer gets in at time **t1** at some station, they get out at time **t2** with **t2 > t1**. All events happen in chronological order.
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**Example 1:**
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```
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Input
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["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
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[[],[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"]]
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Output
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[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]
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Explanation
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UndergroundSystem undergroundSystem = new UndergroundSystem();
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undergroundSystem.checkIn(45, "Leyton", 3);
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undergroundSystem.checkIn(32, "Paradise", 8);
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undergroundSystem.checkIn(27, "Leyton", 10);
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undergroundSystem.checkOut(45, "Waterloo", 15);
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undergroundSystem.checkOut(27, "Waterloo", 20);
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undergroundSystem.checkOut(32, "Cambridge", 22);
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undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
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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
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undergroundSystem.checkIn(10, "Leyton", 24);
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undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.00000
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undergroundSystem.checkOut(10, "Waterloo", 38);
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undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.00000
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```
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**Example 2:**
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```
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Input
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["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
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[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]
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Output
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[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]
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Explanation
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UndergroundSystem undergroundSystem = new UndergroundSystem();
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undergroundSystem.checkIn(10, "Leyton", 3);
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undergroundSystem.checkOut(10, "Paradise", 8);
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undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000
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undergroundSystem.checkIn(5, "Leyton", 10);
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undergroundSystem.checkOut(5, "Paradise", 16);
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undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000
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undergroundSystem.checkIn(2, "Leyton", 21);
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undergroundSystem.checkOut(2, "Paradise", 30);
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undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667
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```
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**Constraints:**
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- There will be at most `20000` operations.
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- `1 <= id, t <= 106`
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- All strings consist of uppercase and lowercase English letters, and digits.
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- `1 <= stationName.length <= 10`
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- Answers within `105` of the actual value will be accepted as correct.
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## 题目大意
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请你实现一个类 UndergroundSystem ，它支持以下 3 种方法：
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- 1. checkIn(int id, string stationName, int t)
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- 编号为 id 的乘客在 t 时刻进入地铁站 stationName 。
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- 一个乘客在同一时间只能在一个地铁站进入或者离开。
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- 2. checkOut(int id, string stationName, int t)
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- 编号为 id 的乘客在 t 时刻离开地铁站 stationName 。
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- 3. getAverageTime(string startStation, string endStation)
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- 返回从地铁站 startStation 到地铁站 endStation 的平均花费时间。
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- 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。
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- 调用 getAverageTime 时，询问的路线至少包含一趟行程。
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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。
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## 解题思路
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- 维护 2 个 `map`。一个 `mapA` 内部存储的是乘客 `id` 与（入站时间，站名）的对应关系。另外一个 `mapB` 存储的是起点站与终点站花费总时间与人数总数的关系。每当有人 `checkin()`，就更新 `mapA` 中的信息。每当有人 `checkout()`，就更新 `mapB` 中的信息，并删除 `mapA` 对应乘客 `id` 的键值对。最后调用 `getAverageTime()` 函数的时候根据 `mapB` 中存储的信息计算即可。
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## 代码
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```go
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package leetcode
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type checkin struct {
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station string
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time int
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}
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type stationTime struct {
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sum, count float64
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}
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type UndergroundSystem struct {
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checkins map[int]*checkin
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stationTimes map[string]map[string]*stationTime
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}
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func Constructor() UndergroundSystem {
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return UndergroundSystem{
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make(map[int]*checkin),
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make(map[string]map[string]*stationTime),
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}
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}
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func (s *UndergroundSystem) CheckIn(id int, stationName string, t int) {
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s.checkins[id] = &checkin{stationName, t}
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}
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func (s *UndergroundSystem) CheckOut(id int, stationName string, t int) {
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checkin := s.checkins[id]
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destination := s.stationTimes[checkin.station]
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if destination == nil {
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s.stationTimes[checkin.station] = make(map[string]*stationTime)
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}
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st := s.stationTimes[checkin.station][stationName]
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if st == nil {
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st = new(stationTime)
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s.stationTimes[checkin.station][stationName] = st
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}
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st.sum += float64(t - checkin.time)
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st.count++
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delete(s.checkins, id)
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}
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func (s *UndergroundSystem) GetAverageTime(startStation string, endStation string) float64 {
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st := s.stationTimes[startStation][endStation]
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return st.sum / st.count
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}
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/**
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* Your UndergroundSystem object will be instantiated and called as such:
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* obj := Constructor();
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* obj.CheckIn(id,stationName,t);
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* obj.CheckOut(id,stationName,t);
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* param_3 := obj.GetAverageTime(startStation,endStation);
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*/
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``` |