1396-设计地铁系统(Design Underground System)
发表于：2021-12-03 | 分类： 中等
字数统计: 2k | 阅读时长: 11分钟 | 阅读量：

原文链接: https://leetcode-cn.com/problems/design-underground-system

英文原文

An underground railway system is keeping track of customer travel times between different stations. They are using this data to calculate the average time it takes to travel from one station to another.

Implement the UndergroundSystem class:

  • void checkIn(int id, string stationName, int t) 
    <ul>
    <li>A customer with a card ID equal to <code>id</code>, checks in at the station <code>stationName</code> at time <code>t</code>.</li>
    <li>A customer can only be checked into one place at a time.</li>
</ul>
</li>
<li><code>void checkOut(int id, string stationName, int t)</code>
<ul>
    <li>A customer with a card ID equal to <code>id</code>, checks out from the station <code>stationName</code> at time <code>t</code>.</li>
</ul>
</li>
<li><code>double getAverageTime(string startStation, string endStation)</code>
<ul>
    <li>Returns the average time it takes to travel from <code>startStation</code> to <code>endStation</code>.</li>
    <li>The average time is computed from all the previous traveling times from <code>startStation</code> to <code>endStation</code> that happened <strong>directly</strong>, meaning a check in at <code>startStation</code> followed by a check out from <code>endStation</code>.</li>
    <li>The time it takes to travel from <code>startStation</code> to <code>endStation</code> <strong>may be different</strong> from the time it takes to travel from <code>endStation</code> to <code>startStation</code>.</li>
    <li>There will be at least one customer that has traveled from <code>startStation</code> to <code>endStation</code> before <code>getAverageTime</code> is called.</li>
</ul>
</li>

You may assume all calls to the checkIn and checkOut methods are consistent. If a customer checks in at time t1 then checks out at time t2, then t1 < t2. 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);  // Customer 45 "Leyton" -> "Waterloo" in 15-3 = 12
undergroundSystem.checkOut(27, "Waterloo", 20);  // Customer 27 "Leyton" -> "Waterloo" in 20-10 = 10
undergroundSystem.checkOut(32, "Cambridge", 22); // Customer 32 "Paradise" -> "Cambridge" in 22-8 = 14
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. One trip "Paradise" -> "Cambridge", (14) / 1 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 11.00000. Two trips "Leyton" -> "Waterloo", (10 + 12) / 2 = 11
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38);  // Customer 10 "Leyton" -> "Waterloo" in 38-24 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 12.00000. Three trips "Leyton" -> "Waterloo", (10 + 12 + 14) / 3 = 12

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); // Customer 10 "Leyton" -> "Paradise" in 8-3 = 5
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000, (5) / 1 = 5
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16); // Customer 5 "Leyton" -> "Paradise" in 16-10 = 6
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000, (5 + 6) / 2 = 5.5
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30); // Customer 2 "Leyton" -> "Paradise" in 30-21 = 9
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667, (5 + 6 + 9) / 3 = 6.66667

 

Constraints:

  • 1 <= id, t <= 106
  • 1 <= stationName.length, startStation.length, endStation.length <= 10
  • All strings consist of uppercase and lowercase English letters and digits.
  • There will be at most 2 * 104 calls in total to checkIn, checkOut, and getAverageTime.
  • Answers within 10-5 of the actual value will be accepted.

中文题目

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。

通过代码

高赞题解

思路

  1. 使用 checkRecord 根据 id 来保存上车 车站名字 和 时间
  2. 在下车车站，把乘车时间计算出来，按两个 车站名字 保存起来，同时还要保存 乘车次数
  3. 取得平均时间时，只需要查询两个车站，得到总时间，和总次数

答题

[]
class UndergroundSystem {
public:
    UndergroundSystem() {}

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

    void checkOut(int id, string stationName, int t) {
        string name = getStationName(checkRecord[id].first, stationName);
        t -= checkRecord[id].second;
        count[name].first += (double)t;
        count[name].second += 1;
    }

    double getAverageTime(string startStation, string endStation) {
        string name = getStationName(startStation, endStation);
        double ans = count[name].first / (double)count[name].second;
        return ans;
    }

private:
    string getStationName(string startStation, string endStation)
    {
        return startStation + "," + endStation;
    }

    unordered_map<int, pair<string, int>> checkRecord;
    unordered_map<string, pair<double, int>> count;
};

致谢

感谢您的观看，希望对您有帮助，欢迎热烈的交流！

如果感觉还不错就点个赞吧~

这是 我的leetcode ，帮助我收集整理题目，可以方便的 visual studio 调试，欢迎关注，star

统计信息

通过次数 提交次数 AC比率
11774 28876 40.8%

提交历史

提交时间 提交结果 执行时间 内存消耗 语言
字符串 哈希表 设计
上一篇:
1395-统计作战单位数(Count Number of Teams)
下一篇:
1397-找到所有好字符串(Find All Good Strings)
本文目录
本文目录
    •
    本站总访问量： 次 |
    Copyright © 2020 - 2021 枫奇丶宛南 | Powered by Hexo | Theme Bamboo