问题描述
给出两个路径,一个父文件夹和一个子文件夹(或孙子文件夹,孙子孙文件夹等),从父文件夹中找到子文件夹的相对深度。始终保证子文件夹始终存在于父文件夹中,并且路径始终以斜杠结尾。
示例:
parent = "/home/user/path/parent/"
child = "/home/user/path/parent/g/g/child/"
#ans = 3
解决方法
解决此问题的最简单方法如下:
parent = '/home/user/path/parent/'
child = '/home/user/path/parent/g/g/child/'
# Split the paths
child_chunks = child.split(File::SEPARATOR).reject(&:empty?)
parent_chunks = parent.split(File::SEPARATOR).reject(&:empty?)
# Find the diff between the two lists
puts (child_chunks - parent_chunks).length # => 3
最好的方法是使用Pathname
助手:
require 'pathname'
parent = Pathname.new('/home/user/path/parent/')
child = Pathname.new('/home/user/path/parent/g/g/child/')
puts child.relative_path_from(parent).to_s.split(File::SEPARATOR).length # => 3
,
def relative_depth(path1,path2)
path2.count(File::SEPARATOR) - path1.count(File::SEPARATOR)
end
path1 = 'a/b/c/'
path2 = 'a/b/c/d/e/f/'
relative_depth(path1,path2)
#=> 3
path3 = 'C:a\\b\\c\\'
path4 = 'C:a\\b\\c\\d\\e\\f\\'
File::SEPARATOR = "\\" # simulate Windows
relative_depth(path3,path4)
#=> 3
请参见String#count。