1396. Design Underground System

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.

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设计一个地铁记录系统。完整代码如下，使用system记录每条线路的用时和乘坐次数，使用starts记录每个人旅行起点。不用记录终点，因为checkout的时候，把起点取出计算用时即可。

class UndergroundSystem {
private:
	map<string, map<string, pair<int, int>>> system_; // start->end, total time, #travels
	map<int, pair<string, int>> starts_; // travel start station
public:
	UndergroundSystem() {

	}

	void checkIn(int id, string stationName, int t) {
		starts_[id] = make_pair(stationName, t);
	}

	void checkOut(int id, string stationName, int t) {
		string start_station = starts_[id].first;
		int start_time = starts_[id].second;
		starts_.erase(id);
		if (system_.find(start_station) == system_.end()
			|| system_[start_station].find(stationName) == system_[start_station].end()) {
			system_[start_station][stationName] = make_pair(0, 0);
		}
		system_[start_station][stationName].first += (t - start_time);
		system_[start_station][stationName].second += 1;
	}

	double getAverageTime(string startStation, string endStation) {
		return system_[startStation][endStation].first / double(system_[startStation][endStation].second);
	}
};

本代码提交AC，用时744MS。

5368. Find Lucky Integer in an Array

Given an array of integers arr, a lucky integer is an integer which has a frequency in the array equal to its value.

Return a lucky integer in the array. If there are multiple lucky integers return the largest of them. If there is no lucky integer return -1.

Example 1:

Input: arr = [2,2,3,4]
Output: 2
Explanation: The only lucky number in the array is 2 because frequency[2] == 2.

Example 2:

Input: arr = [1,2,2,3,3,3]
Output: 3
Explanation: 1, 2 and 3 are all lucky numbers, return the largest of them.

Example 3:

Input: arr = [2,2,2,3,3]
Output: -1
Explanation: There are no lucky numbers in the array.

Example 4:

Input: arr = [5]
Output: -1

Example 5:

Input: arr = [7,7,7,7,7,7,7]
Output: 7

Constraints:

• 1 <= arr.length <= 500
• 1 <= arr[i] <= 500

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给定一个数组，定义幸运数字是它的值与它在数组中出现的次数相等的数，问数组中最大的幸运数字是哪个。

简单题，用一个map计数即可，代码如下：

class Solution {
public:
	int findLucky(vector<int>& arr) {
		map<int, int> lucky;
		for (int i = 0; i < arr.size(); ++i) {
			++lucky[arr[i]];
		}
		int max_lucky = -1;
		for (map<int, int>::iterator it = lucky.begin(); it != lucky.end(); ++it) {
			if (it->first == it->second) {
				max_lucky = max(max_lucky, it->first);
			}
		}
		return max_lucky;
	}
};

本代码提交AC，用时12MS。