一种更优雅的方式来迭代列表以比较相邻的两个元素

Question

我有一个以这种方式工作的方法：

1. 以 3 个参数作为参数 - 包含日期（按升序排序）、间隔单位和间隔值的列表

2. 检查下一个元素是否不超过前一个日期（间隔）。换句话说，给定 30 分钟的间隔，上一个 - 10:00，下一个 10:29 - 进一步迭代。如果下一个是 10:31 - 打破它并返回连续日期的计数器。

它的代码如下：

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，该代码对我来说看起来很奇怪。有什么方法可以让它看起来更具可读性吗？

Answer 1

如果您使用的是 Java 8 或更高版本，则可以改用 java.time-API。它对“时间段”的内置支持使实际实施变得更加简单。static int daysInARow(List<Instant> allInstants, Duration maxDifference) {&nbsp; &nbsp; &nbsp; &nbsp; int counter = allInstants.size() > 0 ? 1 : 0;&nbsp; &nbsp; &nbsp; &nbsp; Instant previous = allInstants.get(0);&nbsp; &nbsp; &nbsp; &nbsp; for (int i = 1; i < allInstants.size(); i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Instant current = allInstants.get(i);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (Duration.between(previous, current).compareTo(maxDifference) > 0)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; counter++;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; previous = current;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return counter;&nbsp; &nbsp; }java.util.Date如果您在项目的其他部分使用，您可以Instant使用以下命令轻松地在 s之间进行转换Date#from(Instant)和Date#toInstant()