Question

我有一系列字符，如下所示：

chars = ["x", "o", "o", "x", "x", "o", "o", "x", "x", "x", "o", "o"]

我需要获取连续字符的数量和该字符串的索引。它应该是这样的：

[
  { index: 0, length: 1 },  # "x"
  { index: 1, length: 2 },  # "o", "o"
  { index: 3, length: 2 },  # "x", "x"
  { index: 5, length: 2 },  # "o", "o"
  { index: 7, length: 3 },  # "x", "x", "x"
  { index: 10, length: 2 }  # "o", "o"
]

有没有简单的方法来实现这一目标？

Answer 1

不确定你是否称之为一种简单的方式，但这是一种单行的方式。结果数组的格式为[index, length]。

chars.each_with_index.chunk {|i, _| i}.map {|_, y| [y.first.last, y.length]}
#=> [[0, 1], [1, 2], [3, 2], [5, 2], [7, 3], [10, 2]]

Answer 2

另外两种方法：

使用Emumerable＃slice_when（v.2.2 +）

count = 0
arr = chars.slice_when { |a,b| a != b }.map do |arr|
  sz = arr.size
  h = { index: count, length: sz }
  count += sz  
  h
end
  #=> [{:index=>0, :length=>1}, {:index=>1, :length=>2}, {:index=>3, :length=>2},
  #    {:index=>5, :length=>2}, {:index=>7, :length=>3}, {:index=>10, :length=>2}]

使用带反向引用的正则表达式

count = 0
arr = chars.join.scan(/((.)\2*)/).map do |run, _|
  sz = run.size
  h = { index: count, length: sz }
  count += sz
  h
end
  #=> [{:index=>0, :length=>1}, {:index=>1, :length=>2}, {:index=>3, :length=>2},
  #    {:index=>5, :length=>2}, {:index=>7, :length=>3}, {:index=>10, :length=>2}]

将数组拆分为子数组

2 个答案: