Question

我正在努力寻找一种有效而清晰的算法来解决以下问题。我特意在javascript中工作，但这确实是一个非常普遍的问题。

我每天都有一家商店开放数小时太阳：关闭星期一：7a-6p 周二：7a-6p ...

每天可以有不同的营业时间，有些日子可能完全关闭。

如果此商店的工作需要一定的时间，并且只在其工作时间内有效。我怎样才能计算出最早的时间？一些考虑因素，一些工作可能需要超过一周的工作时间来完成，商店可能会选择在更随机的时间关闭，6：03。

Answer 1

也许是这样的？你需要有很长的工作才能导致任何性能问题。

var timeToFinish = getTimeToFinish(job);
var timeLeft = timeToFinish;
var date = getStartingDate(job);

while(timeLeft > 0){

    var workingHours = getWorkingHours(day);

    if(timeLeft > workingHours){
        timeLeft -= workingHours;
        date = nextDay(date);
    } else {
        timeLeft == 0;
    }

}

return date;

函数getWorkingHours(day)可能会查询具有开放时间表的数据库。

Answer 2

/**
 * @param hoursOfWork a non-negative number of hours required to complete
 *     a project.
 * @param openHours A list of length 7 of lists of ranges of
 *     inclusive hours in the range [0-23].
 *     For a particular day of the week,
 *     open 9-5 is represented by [[9,17]],
 *     and closed-for-the-day is represented by [],
 *     and open 9-12 and 1-5 is represented by [[9, 12], [13, 17]].
 * @param startDayOfWeek is an integer in the range [0, 6] of the first
 *     day of the week on which work occurs.
 */
function daysRequired(hoursOfWork, startDayOfWeek, openHours) {
  var daysInWeek = 7;  // Vive la revolution (but not metric weeks).

  // Make sure they're open sometime during the week so we don't infinitely loop.
  var totalForWeek = 0;
  var totalsForDay = [];
  for (var i = 0; i < daysInWeek; ++i) {
    var hoursOfWorkInDay = hoursOfWork[i];
    var totalForDay = 0;
    for (var j = 0, n = hoursOfWorkInDay.length; j < n; ++j) {
      totalForDay += (hoursOfWorkInDay[j][1] - hoursOfWorkInDay[j][0]);
    }
    totalForWeek += totalForDay;
    totalsForDay[i] = totalForDay;
  }
  if (totalForWeek < 1) { throw new Error('work harder'); }
  if (hoursOfWork !== hoursOfWork || !isFinite(hoursOfWork)) {
    throw new Error('' + hoursOfWork);
  }

  // Step forward by days subtracting each days work hours from the total.
  var daysRequired = 0;
  var dayOfWeek = startDayOfWeek;
  while (hoursOfWork > 0) {
    ++daysRequired;  // Count the day since we have to do some work on it.
    var totalForDay = totalsForDay[dayOfWeek];
    if (totalForDay >= hoursOfWork) { break; }
    hoursOfWork -= totalForDay;  // Work that day.
    dayOfWeek = (dayOfWeek + 1) % daysInWeek;  // Skip to next day of week.
  }
  return daysRequired;
}

在几小时的范围内寻找时间

2 个答案: