CodeWars一些积累No.1 超市排队的实现

最新推荐文章于 2022-04-06 20:45:07 发布

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本文探讨了一个常见的超市排队问题，即如何优化多窗口的服务时间，以减少所有顾客的总等待时间。通过使用C++的vector和algorithm库，实现了一种有效的算法来解决这一问题。该算法首先初始化n个窗口的时间为0，然后遍历顾客队列，将每个顾客分配到当前服务时间最短的窗口，最后返回所有窗口中服务时间最长的那个窗口的时间。

超市排队问题(多窗口服务时间)

要求

There is a queue for the self-checkout tills at the supermarket. Your task is write a function to calculate the total time required for all the customers to check out!

input

customers: an array of positive integers representing the queue. Each integer represents a customer, and its value is the amount of time they require to check out.
n: a positive integer, the number of checkout tills.

output

The function should return an integer, the total time required.

样例

queueTime(std::vector<int>{5,3,4}, 1)
// should return 12
// because when n=1, the total time is just the sum of the times

queueTime(std::vector<int>{10,2,3,3}, 2)
// should return 10
// because here n=2 and the 2nd, 3rd, and 4th people in the 
// queue finish before the 1st person has finished.

queueTime(std::vector<int>{2,3,10}, 2)
// should return 12

我的实现方式

#include <algorithm>
#include <vector>

long queueTime(std::vector<int> customers,int n){
    std::vector<int> windows(n,0);            //用向量表示n个窗口
    int Num_Customers = customers.size();    //存储顾客的数量
    for ( int i = 0; i < Num_Customers ; i++ )
    {
        int position = min_element(windows.begin(),windows.end()) - windows.begin(); 
        //找到最小窗口的对应位置
        windows[position] += customers[i];
    }
    long MaxValue = *max_element(windows.begin(),windows.end());
    return MaxValue;