Question

我正在编写一个程序，用于在ArrayList中存储约会。要求用户键入约会的日期，开始时间和结束时间。程序需要在用户想要提交新约会时检查是否已经有约会。这是递归二进制搜索部分。这是我到目前为止实施的内容。

方法：

public boolean overlap(ArrayList<Appointment> list, int start, int end, 
                       Calendar startCal, Calendar endCal)
{
    int middle = (start + end)/2;
    System.out.println("Start: " + start + " End: " + end + " Middle: " + middle);

    Appointment appt = list.get(middle);
    if(start == end){
        return((startCal.compareTo(appt.getStart()) >= 0 && startCal.compareTo(appt.getEnd()) <= 0 ) ||
                endCal.compareTo(appt.getEnd()) <= 0 && endCal.compareTo(appt.getStart()) >= 0 );
    }
    else if(startCal.compareTo(appt.getStart()) <= 0 && endCal.compareTo(appt.getStart()) <= 0){
        return overlap(list, start, middle - 1, startCal, endCal);
    }
    else{
        return overlap(list, middle, end, startCal, endCal); 
    }  
}

我如何称呼它：

case 1 : //Enter a new appointment
                setCal();
                startDate = new GregorianCalendar(year, month, day, hour, minute);
                endDate = new GregorianCalendar(year, month, day, endHour, endMinute);

                appt = new Appointment(description, startDate, endDate);

                if(apptList.isEmpty()){
                    apptList.add(appt);
                }
                else{
                    boolean isOverlap = overlap(apptList, 0, apptList.size() - 1, 
                                        appt.getStart(), appt.getEnd());

                    if(isOverlap){
                        System.out.println("");
                        System.out.println("ERROR: There is already an appointment at that time");
                        System.out.println("");
                    }
                    else{
                        int indexToBeAdded = indexFinder(apptList, 0, apptList.size() -1, 
                                                       appt.getStart(), appt.getEnd());
                        Appointment tmp = apptList.get(indexToBeAdded);

                        if(tmp.start.compareTo(appt.start) < 0){
                            apptList.add(indexToBeAdded + 1, appt);   
                        }
                        else{
                            apptList.add(indexToBeAdded , appt);
                        }
                    }
                }
                choice = 0;
                break;

setCal（）只是提示用户输入所有日历参数，并根据输入的内容设置它们。另外，indexFinder是另一种递归二进制搜索方法，它找到应该添加约会的索引，以便添加约会正确的订单。

截至目前，在我第三次约会时，我尝试添加一个堆栈溢出异常。

感谢您的帮助

Answer 1

return overlap(list, start, middle - 1, startCal, endCal);

如果你用start = 0和end = 1调用重叠，中间= 0 + 1/2 = 0，那么你用start = 0和end = -1调用重叠...这意味着它们会进一步飞行进一步分开，你跳过了start == end的停止条件。

尝试将start == end转为start > end。

编辑：问题二是这个电话

return overlap(list, middle, end, startCal, endCal);

如果用start = 0和end = 1调用重叠，中间= 0 + 1/2 = 0，那么你用start = 0和end = 1调用重叠...无限递归循环！堆栈溢出！它永远不会结束。

要解决此问题，请将middle转换为middle + 1。

执行递归二进制搜索

1 个答案: