Question

我正在尝试创建一个输入两个字符串的程序。如果它们匹配，则返回它们包含相同长度2子串的位置数。

例如：string_match（'xxcaazz'，'xxbaaz'）→3“xx”“aa”“az” 我的问题是我应该使用哪些元字符来验证

这是我的想法

puts "enter word"
a = STDIN.gets
a.chomp!
puts"enter word"
b = STDIN.gets
b.chomp!
if a == /word/ or b == /word/ then 
  puts str.match(/{a} {b}/) + "equal"
end

Answer 1

更新的答案：

（仍然飙升，但更好）

first_word = 'xxcaazz'.split('')
second_word ='xxbaaz'.split('')

first_word_length = first_word.length
second_word_length = second_word.length

if [first_word_length, second_word_length].min == first_word_length
  inner_word = second_word
  outter_word = first_word
else
  inner_word = first_word
  outter_word = second_word
end

outter_word_length = outter_word.length - 2

word_matches = 0

(0..outter_word_length).each do |character|
  if "#{outter_word[character]}#{outter_word[character + 1]}" == "#{inner_word[character]}#{inner_word[character + 1]}"
    puts "#{outter_word[character]}#{outter_word[character + 1]}"
    word_matches += 1
  end
end

puts "Found #{word_matches} matches"

原始SPIKE：

这可能会让你有一个良好的开端（虽然它绝不是防弹，只是一个快速的飙升）：

first_word = 'xxcaazz'.split('')
second_word ='xxbaaz'.split('')
first_word_length = first_word.length

(0..first_word_length).each do |character|
  if "#{second_word[character]}#{second_word[character + 1]}" == "#{first_word[character]}#{first_word[character + 1]}"
    puts "#{second_word[character]}#{second_word[character + 1]}"
  end
end

Answer 2

解包'a2X'表示提取2个字节，然后倒回1个字节：

first_word = 'xxcaazz'
second_word ='xxbaaz'
tokens = first_word.unpack 'a2X' * (first_word.length - 1)
# => ["xx", "xc", "ca", "aa", "az", "zz"]
tokens.flat_map{|m| second_word.scan m}
# => ["xx", "aa", "az"]

具有匹配字符串的ruby元字符

2 个答案: