我有一個以這種方式工作的方法：以 3 個參數(shù)作為參數(shù) - 包含日期（按升序排序）、間隔單位和間隔值的列表檢查下一個元素是否不超過前一個日期（間隔）。換句話說，給定 30 分鐘的間隔，上一個 - 10:00，下一個 10:29 - 進一步迭代。如果下一個是 10:31 - 打破它并返回連續(xù)日期的計數(shù)器。它的代碼如下：public static void main(String[] args){    Date d1 = new Date();    Date d2 = addOrSubtractTimeUnitFromDate(d1, Calendar.MINUTE, 10, true);    Date d3 = addOrSubtractTimeUnitFromDate(d2, Calendar.MINUTE, 10, true);    Date d4 = addOrSubtractTimeUnitFromDate(d3, Calendar.MINUTE, 10, true);    Date d5 = addOrSubtractTimeUnitFromDate(d4, Calendar.MINUTE, 10, true);    Date d6 = addOrSubtractTimeUnitFromDate(d5, Calendar.MINUTE, 10, true);    List<Date> threeDates = new ArrayList<>();    threeDates.add(d1);    threeDates.add(d2);    threeDates.add(d3);    threeDates.add(d4);    threeDates.add(d5);    threeDates.add(d6);    System.out.println(returnDatesInARowCounter(threeDates, Calendar.MINUTE, 30));}private static int returnDatesInARowCounter(List<Date> allDates, int intervalBetween2DatesTimeUnit, int intervalValue){    int datesInARowCounter = allDates.size() > 0 ? 1 : 0; // esp. this line (in case allDates is empty)    Date lastDate = null;    Date nextDate;    Iterator<Date> iter = allDates.iterator();    while (iter.hasNext())    {        nextDate = iter.next();        if (lastDate != null) // both lastDate и nextDate are initialized now        {            if(isNextIncidentInIntervalWithLastOneOrNot(lastDate, nextDate, intervalBetween2DatesTimeUnit, intervalValue, true))            {                datesInARowCounter += 1;            }            else break;        }        lastDate = nextDate;    }    return datesInARowCounter;}public static Date addOrSubtractTimeUnitFromDate(Date dateToAddToOrSubtractFrom, int calendarTimeUnit, int value, boolean isAdd){    if(!isAdd)    {        value = -value;    }    Calendar cal = Calendar.getInstance();    cal.setTime(dateToAddToOrSubtractFrom);    cal.add(calendarTimeUnit, value);    return cal.getTime();}然而，該代碼對我來說看起來很奇怪。有什么方法可以讓它看起來更具可讀性嗎？