Question

我有一个数组数组：

a = [[1, 2, 3], [3, 4, 5], [6, 7, 8], [8, 9], [9, 10]]

我想合并包含一个或多个相同值的所有数组。所以：

a = [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]

我正在努力为此找到一个简洁的方法。有什么想法吗？

Answer 1

我认为这是正确的：

def merge_em(a)
  return a if a.empty?
  rest = a.dup
  groups = []
  group = []
  while rest.any?
    group = rest.shift if group.empty?
    if i = rest.each_index.find { |i| (rest[i] & group).any? }
      group |= rest[i]
      rest.delete_at(i) 
      groups << group if rest.empty?
    else
      groups << group
      group = []
    end
  end
  groups
end

a = [[1, 2, 3], [3, 4, 5], [98, 99], [6, 7, 8], [8, 9], [9, 10]]
merge_em(a)
  #=> [[1, 2, 3, 4, 5], [98, 99], [6, 7, 8, 9, 10]]

Answer 2

这是怎么回事：

a = [[1, 2, 3], [3, 4, 5], [98, 99], [6, 7, 8], [8, 9], [9, 10]]

b = [ a.shift ] # start with first sequence
a.each do |seq|
  match = nil
  b.each do |existing_sequence|
    match = existing_sequence if existing_sequence.any? {|x| seq.include? x }
  end
  if match
    seq.each {|x| match << x } # coalese matching fragment
  else
    b << seq # new fragment
  end
end

p b.map {|x| x.uniq }

# => [[1, 2, 3, 4, 5], [98, 99], [6, 7, 8, 9, 10]]

合并数组，如果它们包含一个或多个相同的值

2 个答案: