Question

我正在尝试编写一种方法，其中客户必须与出纳员一起花费一定的时间。

如果第二个数量已过期，则应该返回出纳员是空闲的，因此返回布尔值true。如果时间尚未过去，则应该返回false，因此柜员很忙。但我不确定该怎么做。

这些是我的出纳员和客户类我不确定如何编写其中一种方法。

这是Customer类：

import java.util.Random;

public class Customer {
    //create a random number between 2 and 5 seconds it takes to process a customer through teller
    Random number = new Random();

    protected int arrivalTime; //determines the arrival time of a customer to a bank
    protected int processTime; //the amount of time the customer spend with the teller

    //a customer determines the arrival time of a customer
    public Customer(int at) {
        arrivalTime = at;
        processTime = number.nextInt(5 - 2 + 2) + 2; //the process time is between 2 and 5 seconds
    }

    public int getArrival() {
        return arrivalTime;
    }

    public int getProcess() {
        return processTime; //return the processTime
    }

    public boolean isDone() {
        //this is the method I am trying to create once the  process Time is expired
        //it will return true which means the teller is unoccupied

        //if the time is not expired it will return false and teller is occupied
        boolean expired = false;
        return expired;
    }
}

这是Teller类：

public class Teller {
    //variables
    boolean free; //determines if teller if free
    Customer rich; //a Customer object

    public Teller() {
        free = true; //the teller starts out unoccupied 
    }

    public void addCustomer(Customer c) { //a customer is being added to the teller
        rich = c;
        free = false; //the teller is occupied 
    }

    public boolean isFree() {
        if (rich.isDone() == true) {
            return true; //if the customer is finished at the teller the return true
        } else {
            return false;
        }
    }
}

Answer 1

public boolean isDone() {
    long time = System.currentTimeMillis();

    if (Integer.parseInt(time-arrivalTime)/1000 >= processTime) {
        return true;
    }
    return false;
}

试试这个，你必须让你的到达时间到long

将构造函数中的arrivalTime初始化为 arrivalTime = System.currentTimeMillis();

Answer 2

您可以使用System.currentTimeMillis()。

long start = System.currentTimeMillis();
// ...
long current = System.currentTimeMillis();
long elapsed = current - start;

要将其转换为秒，只需除以1000。

如何创建一个布尔方法，确定是否已经通过了第二个

2 个答案: