在Ruby中找到两个字符串之间的第一个不同字符

I have two strings.

我有两个字符串。

str_a = "the_quick_brown_fox"
str_b = "the_quick_red_fox"

I want to find the first index at which the two strings differ (i.e. str_a[i] != str_b[i]).

我想找到两个字符串不同的第一个索引(即str_a[I] != str_b[I])。

I know I could solve this with something like the following:

我知道我可以用以下方法来解决这个问题:

def diff_char_index(str_a, str_b)
  arr_a, arr_b = str_a.split(""), str_b.split("")
  return -1 unless valid_string?(str_a) && valid_string?(str_b)
  arr_a.each_index do |i|
    return i unless arr_a[i] == arr_b[i]
  end
end

def valid_string?(str)
  return false unless str.is_a?(String)
  return false unless str.size > 0
  true
end

diff_char_index(str_a, str_b) # => 10

Is there a better way to do this?

有更好的方法吗?

4 个解决方案

#1

Something like this ought to work:

像这样的东西应该有用:

str_a.each_char.with_index
  .find_index {|char, idx| char != str_b[idx] } || str_a.size

Edit: It works: http://ideone.com/Ttwu1x
Edit 2: My original code returned nil if str_a was shorter than str_b. I've updated it to work correctly (it will return str_a.size, so if e.g. the last index in str_a is 3, it will return 4).

编辑:它工作:http://ideone.com/Ttwu1x编辑2:如果str_a小于str_b，我的原始代码返回nil。我已经更新它以正确工作(它将返回str_a。大小，例如，如果str_a中的最后一个索引是3，它将返回4)。

Here's another method that may strike some as slightly simpler:

还有一种方法可能会让一些人觉得有点简单:

(0...str_a.size).find {|i| str_a[i] != str_b[i] } || str_a.size

http://ideone.com/275cEU

#2

i = 0
i += 1 while str_a[i] and str_a[i] == str_b[i]
i

#3

str_a = "the_quick_brown_dog"
str_b = "the_quick_red_dog"

(0..(1.0)/0).find { |i| (str_a[i] != str_b[i]) || str_a[i].nil? }
  #=> 10

str_a = "the_quick_brown_dog"
str_b = "the_quick_brown_dog"

(0..(1.0)/0).find { |i| (str_a[i] != str_b[i]) || str_a[i].nil? }
  #=> 19
str_a.size
  #=> 19

#4

This uses a binary search to find the index where a slice of str_a no longer occurs at the beginning of str_b:

这使用了一个二分查找来查找在str_b开始时不再出现的str_a片的索引:

(0..str_a.length).bsearch { |i| str_b.rindex(str_a[0..i]) != 0 }

#1