ruby-on-rails - 如何组合重叠的时间范围(时间范围联合)

Question

我有一个包含多个时间范围的数组:

[Tue, 24 May 2011 08:00:00 CEST +02:00..Tue, 24 May 2011 13:00:00 CEST +02:00,
 Tue, 24 May 2011 16:30:00 CEST +02:00..Tue, 24 May 2011 18:00:00 CEST +02:00,
 Tue, 24 May 2011 08:00:00 CEST +02:00..Tue, 24 May 2011 09:00:00 CEST +02:00,
 Tue, 24 May 2011 15:30:00 CEST +02:00..Tue, 24 May 2011 18:00:00 CEST +02:00]

我想获得具有重叠时间范围组合的相同数组，因此这种情况的输出将是:

[Tue, 24 May 2011 08:00:00 CEST +02:00..Tue, 24 May 2011 13:00:00 CEST +02:00,
 Tue, 24 May 2011 15:30:00 CEST +02:00..Tue, 24 May 2011 18:00:00 CEST +02:00]

因此，当时间范围重叠时，它会创建一个新的时间范围，依此类推。如果它们不重叠，则将保持分开。另一个例子:

输入:

[Tue, 24 May 2011 08:00:00 CEST +02:00..Tue, 24 May 2011 13:00:00 CEST +02:00,
 Tue, 24 May 2011 16:00:00 CEST +02:00..Tue, 24 May 2011 18:00:00 CEST +02:00]

输出(将相同，因为它们不重叠):

[Tue, 24 May 2011 08:00:00 CEST +02:00..Tue, 24 May 2011 13:00:00 CEST +02:00,
 Tue, 24 May 2011 16:00:00 CEST +02:00..Tue, 24 May 2011 18:00:00 CEST +02:00]

我正在考虑某种递归方法，但我需要一些指导......

Answer 1

给定一个函数，如果两个范围重叠则返回 truthy:

def ranges_overlap?(a, b)
  a.include?(b.begin) || b.include?(a.begin)
end

(此功能由 sepp2k and steenslag 提供)

以及合并两个重叠范围的函数:

def merge_ranges(a, b)
  [a.begin, b.begin].min..[a.end, b.end].max
end

然后这个函数，给定一个范围数组，返回一个合并了所有重叠范围的新数组:

def merge_overlapping_ranges(overlapping_ranges)
  overlapping_ranges.sort_by(&:begin).inject([]) do |ranges, range|
    if !ranges.empty? && ranges_overlap?(ranges.last, range)
      ranges[0...-1] + [merge_ranges(ranges.last, range)]
    else
      ranges + [range]
    end
  end
end