递归二分搜索的实现[关闭]

Question

我正在编写一个将约会存储在 ArrayList 中的程序。要求用户输入约会的日期、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。这是我目前已经实现的。

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); 
    }  
}

我怎么称呼它：

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他们输入的内容进行设置。此外，indexFinder 是另一种递归二进制搜索方法，它查找应添加约会的索引，以便将约会添加到正确的顺序。

截至第三次约会时，我尝试添加时遇到堆栈溢出异常。

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，middle = 0+1/2 = 0 来调用重叠，那么你用 start = 0 和 end = -1 来调用重叠......这意味着它们会飞得更远并且距离更远，您跳过了到达start == end 的停止条件。

尝试将start == end 转换为start > end。

编辑：问题二是这个电话

return overlap(list, middle, end, 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......无限递归循环！堆栈溢出！它永远不会结束。

要解决这个问题，请将middle 变成middle + 1。