java - 如何有效地计算不包括时间 block 的时间之间的距离？

Question

我正在尝试计算两个日期之间的分钟数，同时排除任意定义且每周发生的时间段。我还需要能够反向计算，在给定时间的情况下，计算向前的 X 分钟数，不包括这些时间段。

例如，我可能有两个时段 [Fri 5:31pm - Sat 2:26pm] 和 [Tuesday 3:37am - Thursday 1:14am] 在计算两个日期之间的分钟数时我不想计算并在向前计算时。

我目前的代码只针对一个间隙执行此操作，但它不是非常高效并且正在对我的系统造成压力。我还需要适应我目前没有做的多个定义的差距。

我为一个间隙执行此操作的代码如下所示(hideStart 和 hideEnter 是间隙的开始和结束 DateTime，absoluteLowValue是我计算距离或时间的起始时间):

public int absoluteDistance(DateTime high){
    long totalMinutes = new Duration(absoluteLowValue,high).getStandardMinutes();

    if (!gapHider.isHidingGaps())
        return (int)totalMinutes;

    int minutesPerWeek = 10080;
    long minutesPerHide = new Duration(hideStart, hideEnd).getStandardMinutes();

    long numFullWeeks = totalMinutes/minutesPerWeek;
    long remainder = totalMinutes%minutesPerWeek;

    totalMinutes -= numFullWeeks*minutesPerHide;

    DateTime latestEnd = high;
    if (latestEnd.getDayOfWeek() == hideEnd.getDayOfWeek() && latestEnd.getSecondOfDay() < hideEnd.getSecondOfDay()){
        latestEnd = latestEnd.minusWeeks(1);
    }
    while (latestEnd.getDayOfWeek() != hideEnd.getDayOfWeek())
        latestEnd = latestEnd.minusDays(1);
    latestEnd = latestEnd.withTime(hideEnd.getHourOfDay(),
                                hideEnd.getMinuteOfHour(),
                                hideEnd.getSecondOfMinute(),
                                hideEnd.getMillisOfSecond());

    DateTime latestStart = high;
    if (latestStart.getDayOfWeek() == hideStart.getDayOfWeek() && latestStart.getSecondOfDay() < hideStart.getSecondOfDay()){
        latestStart = latestStart.minusWeeks(1);
    }
    while (latestStart.getDayOfWeek() != hideStart.getDayOfWeek())
        latestStart = latestStart.minusDays(1);
    latestStart = latestStart.withTime(hideStart.getHourOfDay(),
                                    hideStart.getMinuteOfHour(),
                                    hideStart.getSecondOfMinute(),
                                    hideStart.getMillisOfSecond());

    long timeToNearestEnd = new Duration(latestEnd, high).getStandardMinutes();
    long timeToNearestStart = new Duration(latestStart, high).getStandardMinutes();

    if (timeToNearestEnd < remainder){
        totalMinutes -= minutesPerHide;
    }else if (timeToNearestStart < remainder){
        totalMinutes -= new Duration(latestStart, high).getStandardMinutes();
    }

    return (int)totalMinutes;
}

public DateTime timeSinceAbsLow(int index){ 
    if (absoluteLowValue != null){

        if (!gapHider.isHidingGaps())
            return absoluteLowValue.plusMinutes(index);

        DateTime date = absoluteLowValue;
        long minutesPerWeek = 10080;
        long minutesPerHide = new Duration(hideStart, hideEnd).getStandardMinutes();
        int difference = (int)(minutesPerWeek - minutesPerHide);
        int count = 0;

        while (index - count >= difference){
            date = date.plusWeeks(1);
            count += difference;
        }

        int remaining = index - count;

        DateTime nextStart = date;

        while (nextStart.getDayOfWeek() != hideStart.getDayOfWeek())
            nextStart = nextStart.plusDays(1);
        nextStart = nextStart.withTime(hideStart.getHourOfDay(),
                                    hideStart.getMinuteOfHour(),
                                    hideStart.getSecondOfMinute(),
                                    hideStart.getMillisOfSecond());

        long timeDiff = new Duration(date, nextStart).getStandardMinutes();

        if (timeDiff < remaining){
            date = nextStart.plusMinutes((int)minutesPerHide);
            count+= timeDiff;
            remaining = index - count;
        }

        date = date.plusMinutes(remaining);
        return date;
    }
    return new DateTime(); 
}

执行此过程是否有更好或更简单的方法？我想，如果我添加大量逻辑来循环遍历“差距”列表，它只会减慢速度。我愿意不使用 Jodatime，我只是碰巧目前正在使用它。任何帮助表示赞赏!

Answer 1

如果我没理解错的话，您想要计算开始日期和结束日期之间的总时间，然后减去一个或多个周期。因此，您可以使用 Duration 对象，然后最后转换为您想要的任何值(分钟、秒等):

// compute all periods you don't want to count
List<Duration> hideList = new ArrayList<>();
// duration between friday and saturday (assuming they have the values of your example)
hideList.add(new Duration(friday, saturday));
// add as many periods you want

// total duration between start and end dates
Duration totalDuration = new Duration(start, end);

// subtract all periods from total
for (Duration duration : hideList) {
    totalDuration = totalDuration.minus(duration);
}

// convert to total number of minutes
long totalMinutes = totalDuration.getStandardMinutes();

我假设 hideList 中使用的所有日期都在开始日期和结束日期之间(但是使用 isAfter 和 isBefore 很容易检查这一点> 方法)。

---

相反，您将 totalMinutes 添加到开始日期，然后将所有持续时间加到 hideList 中:

// sum total minutes to start
DateTime dt = start.plusMinutes((int) totalMinutes);

// sum all the periods
for (Duration duration : hideList) {
    dt = dt.plus(duration);
}
// dt is the end date

有一个细节:当将 Duration 转换为分钟数时，您将失去秒和毫秒的精度，因此当执行相反的算法时，这些字段将会丢失。如果这不是问题，请按上述方法进行。但是，如果您想保留所有精度，只需从添加 totalDuration 对象开始，而不是添加分钟数:

// sum totalDuratin to start (to preserve seconds and milliseconds precision)
DateTime dt = start.plus(totalDuration);