ruby - 在任意时间范围内找到最佳日/月/年间隔的算法？

Question

如果你有一个时间表，说:

2009 年 3 月 19 日 - 2011 年 7 月 15 日

是否有一种算法可以将该时间范围分解为:

March 19, 2009  - March 31, 2009    # complete days
April 1, 2009   - December 31, 2009 # complete months
January 1, 2010 - December 31, 2010 # complete years
January 1, 2011 - June 30, 2011     # complete months
July 1, 2011    - July 15, 2011     # complete days

更具体地说，给定任何任意时间范围，甚至精确到秒，是否有一种算法可以将其分成最佳数量的任意大小间隔？

因此，也许您不想将上述日期范围划分为天/月/年，而是将其划分为 5 天和 18 个月的 block ，类似这样的随机 block 。这种算法有正式名称吗？一个 ruby​​ 示例会很棒，但任何语言都可以。

我已经能够将一些硬编码的 Ruby 东西组合在一起来处理日/月/年示例:

• https://gist.github.com/1165914

...但似乎应该有一种算法来抽象它来处理任何间隔故障。也许它只是归结为简单的数学。

Answer 1

这可能是作弊，但您可以使用 active_support 提供的日期函数大大简化代码。下面是我使用 active_support 想出的一些代码。该算法非常简单。算出第一个月的最后一天，算出最后一个月的第一天。然后打印第一个月，把中间分成几年打印，然后打印最后一个月。当然，由于边缘情况，这个简单的算法在很多方面都失败了。以下算法尝试优雅地解决这些边缘情况中的每一个。希望对您有所帮助。

require 'active_support/all'

# Set the start date and end date
start_date = Date.parse("March 19, 2009")
end_date = Date.parse("July 15, 2011")

if end_date < start_date
  # end date is before start date, we are done
elsif end_date == start_date
  # end date is the same as start date, print it and we are done
  print start_date.strftime("%B %e, %Y")
elsif start_date.year == end_date.year && start_date.month == end_date.month
  # start date and end date are in the same month, print the dates and we
  # are done
  print start_date.strftime("%B %e, %Y"), " - ",
    end_date.strftime("%B %e, %Y"), "\n"
else
  # start date and end date are in different months
  first_day_of_next_month = Date.new((start_date + 1.month).year,
    (start_date + 1.month).month, 1);
  first_day_of_end_month = Date.new(end_date.year, end_date.month, 1);

  # print out the dates of the first month
  if (first_day_of_next_month - 1.day) == start_date
    print start_date.strftime("%B %e, %Y"), "\n"
  else
    print start_date.strftime("%B %e, %Y"), " - ",
      (first_day_of_next_month - 1.day).strftime("%B %e, %Y"), "\n"
  end

  # now print the inbetween dates
  if first_day_of_next_month.year == (first_day_of_end_month - 1.day).year &&
    (first_day_of_end_month - 1.day) > first_day_of_next_month
    # start date and end date are in the same year, just print the inbetween
    # dates 
    print first_day_of_next_month.strftime("%B %e, %Y"), " - ",
      (first_day_of_end_month - 1.day).strftime("%B %e, %Y") + "\n"
  elsif first_day_of_next_month.year < (first_day_of_end_month - 1.day).year
    # start date and end date are in different years so we need to split across
    # years
    year_iter = first_day_of_next_month.year

    # print out the dates from the day after the first month to the end of the
    # year
    print first_day_of_next_month.strftime("%B %e, %Y"), " - ",
      Date.new(first_day_of_next_month.year, 12, 31).strftime("%B %e, %Y"),
      "\n"
    year_iter += 1

    # print out the full intermediate years
    while year_iter < end_date.year
      print Date.new(year_iter, 1, 1).strftime("%B %e, %Y"), " - ",
        Date.new(year_iter, 12, 31).strftime("%B %e, %Y"), "\n"
      year_iter += 1
    end

    # print from the begining of the last year until the last day before the the
    # end month
    print Date.new(first_day_of_end_month.year, 1, 1).strftime("%B %e, %Y"),
      " - ", (first_day_of_end_month - 1.day).strftime("%B %e, %Y"), "\n"
  end

  # finally print out the days of the last month
  if first_day_of_end_month == end_date
    print end_date.strftime("%B %e, %Y"), "\n"
  else
    print first_day_of_end_month.strftime("%B %e, %Y"), " - ",
      end_date.strftime("%B %e, %Y"), "\n"
  end
end