1396. Design Underground System #

题目 #

Implement the UndergroundSystem class:

void checkIn(int id, string stationName, int t)
- A customer with a card id equal to id, gets in the station stationName at time t.
- A customer can only be checked into one place at a time.
void checkOut(int id, string stationName, int t)
- A customer with a card id equal to id, gets out from the station stationName at time t.
double 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. 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.

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 <= 106
All strings consist of uppercase and lowercase English letters, and digits.
1 <= stationName.length <= 10
Answers within 105 of the actual value will be accepted as correct.

题目大意 #

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。

解题思路 #

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。

代码 #

package leetcode

type checkin struct {
	station string
	time    int
}

type stationTime struct {
	sum, count float64
}

type UndergroundSystem struct {
	checkins     map[int]*checkin
	stationTimes map[string]map[string]*stationTime
}

func Constructor() UndergroundSystem {
	return UndergroundSystem{
		make(map[int]*checkin),
		make(map[string]map[string]*stationTime),
	}
}

func (s *UndergroundSystem) CheckIn(id int, stationName string, t int) {
	s.checkins[id] = &checkin{stationName, t}
}

func (s *UndergroundSystem) CheckOut(id int, stationName string, t int) {
	checkin := s.checkins[id]
	destination := s.stationTimes[checkin.station]
	if destination == nil {
		s.stationTimes[checkin.station] = make(map[string]*stationTime)
	}
	st := s.stationTimes[checkin.station][stationName]
	if st == nil {
		st = new(stationTime)
		s.stationTimes[checkin.station][stationName] = st
	}
	st.sum += float64(t - checkin.time)
	st.count++
	delete(s.checkins, id)
}

func (s *UndergroundSystem) GetAverageTime(startStation string, endStation string) float64 {
	st := s.stationTimes[startStation][endStation]
	return st.sum / st.count
}

/**
 * Your UndergroundSystem object will be instantiated and called as such:
 * obj := Constructor();
 * obj.CheckIn(id,stationName,t);
 * obj.CheckOut(id,stationName,t);
 * param_3 := obj.GetAverageTime(startStation,endStation);
 */

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