一種遍歷列表以比較兩個相鄰元素的更優雅的方法

Question

我有一種方法可以這樣工作：

1. 以3個參數作為參數-帶有日期（按升序排列），間隔單位和間隔值的列表
2. 檢查下一個元素是否不超過上一個日期（間隔）。 換句話說，給定間隔30分鍾，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，然后連續返回日期計數器。

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();
}

private static boolean isNextIncidentInIntervalWithLastOneOrNot(Date lastIncidentRegDate, Date nextIncidentRegDate, int intervalTimeUnit, int intervalValue, boolean isBetween)
{
    Date currentIncidentPlusInterval = addOrSubtractTimeUnitFromDate(lastIncidentRegDate, intervalTimeUnit, intervalValue, true);
    boolean betweenBool = isDateBetween(nextIncidentRegDate, lastIncidentRegDate, currentIncidentPlusInterval);

    return isBetween == betweenBool;
}

private static boolean isDateBetween(Date targetDate, Date startDate, Date endDate)
{
    return targetDate.compareTo(startDate) >= 0 && targetDate.compareTo(endDate) <= 0;
}

但是，代碼對我來說似乎是特有的。 有什么辦法使它看起來更具可讀性？

Answer 1

如果您使用的是Java 8或更高版本，則可以改用java.time-API。 它對“時間段”的內置支持使實際實現變得更加簡單。

static int daysInARow(List<Instant> allInstants, Duration maxDifference) {
        int counter = allInstants.size() > 0 ? 1 : 0;
        Instant previous = allInstants.get(0);

        for (int i = 1; i < allInstants.size(); i++) {
            Instant current = allInstants.get(i);
            if (Duration.between(previous, current).compareTo(maxDifference) > 0)
                break;
            counter++;
            previous = current;
        }

        return counter;
    }

如果在項目的其他部分中使用java.util.Date ，則可以通過使用以下命令輕松地在Instant之間進行轉換

Date#from(Instant)

和

Date#toInstant()