ruby – 在任意时间范围内查找最佳日月年间隔的算法？

概述如果您有时间表,请说： 2009年3月19日 – 2011年7月15日 是否有一种算法可以将时间框架分解为： March 19, 2009 - March 31, 2009 # complete daysApril 1, 2009 - December 31, 2009 # complete monthsJanuary 1, 2010 - December 31, 2010 # c 如果您有时间表,请说：

2009年3月19日 – 2011年7月15日

是否有一种算法可以将时间框架分解为：

march 19,2009  - march 31,2009    # complete daysApril 1,2009   - December 31,2009 # complete monthsJanuary 1,2010 - December 31,2010 # complete yearsJanuary 1,2011 - June 30,2011     # complete monthsJuly 1,2011    - July 15,2011     # complete days

更具体地说,给定任意时间帧,甚至到第二个时间帧,是否有一种算法可以将其划分为最大数量的任意大小的间隔？

因此,可能不是将上述日期范围划分为几天/几个月/几年,而是将其划分为5天和18个月的块,这样就是随机的.这种算法有正式的名称吗？一个ruby的例子很棒,但任何语言都可以.

我已经能够破解一些硬编码的Ruby来处理日/月/年的例子：

> https://gist.github.com/1165914

…但似乎应该有一个算法来抽象它来处理任何间隔细分.也许它只是归结为简单的数学.

解决方法 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提出的一些代码.算法非常简单.弄清楚第一个月的最后一天,弄清楚上个月的第一天.然后打印第一个月,将中间分成几年并打印出来,然后打印上个月.当然,由于边缘情况,这种简单算法在许多方面都有所下降.以下算法尝试优雅地处理每个边缘情况.我希望它有所帮助.

require 'active_support/all'# Set the start date and end datestart_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 doneelsif 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,"\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,"\n"  else    print start_date.strftime("%B %e,(first_day_of_next_month - 1.day).strftime("%B %e,"\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,(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,Date.new(first_day_of_next_month.year,12,31).strftime("%B %e,"\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,Date.new(year_iter,"\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,"\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,"\n"  else    print first_day_of_end_month.strftime("%B %e,"\n"  endend

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