我怎么算元音？
我已经看到了解决方案，它或多或少匹配
Write a method that takes a string and returns the number of vowels in the string. You may assume that all the letters are lower cased. You can treat "y" as a consonant. Difficulty: easy. def count_vowels(string) vowel = 0 i = 0 while i < string.length if (string[i]=="a" || string[i]=="e" || string[i]=="i" || string[i]=="o"|| string[i]=="u") vowel +=1 end i +=1 return vowel end puts("count_vowels(\"abcd\") == 1: #{count_vowels("abcd") == 1}") puts("count_vowels(\"color\") == 2: #{count_vowels("color") == 2}") puts("count_vowels(\"colour\") == 3: #{count_vowels("colour") == 3}") puts("count_vowels(\"cecilia\") == 4: #{count_vowels("cecilia") == 4}") def count_vowels(str) str.scan(/[aeoui]/).count end
/[aeoui]/是一个正则表达式，基本上意味着“任何这些字符：a，e，o，u，i”。 String#scan方法返回String#scan则表达式的所有匹配项。
def count_vowels(str) str.count("aeoui") end
你的function很好，你只是错过了关闭你的while循环的关键字
def count_vowels(string) vowel = 0 i = 0 while i < string.length if (string[i]=="a" || string[i]=="e" || string[i]=="i" || string[i]=="o"|| string[i]=="u") vowel +=1 end i +=1 end return vowel end puts("count_vowels(\"abcd\") == 1: #{count_vowels("abcd") == 1}") puts("count_vowels(\"color\") == 2: #{count_vowels("color") == 2}") puts("count_vowels(\"colour\") == 3: #{count_vowels("colour") == 3}") puts("count_vowels(\"cecilia\") == 4: #{count_vowels("cecilia") == 4}") #=> count_vowels("abcd") == 1: true #=> count_vowels("color") == 2: true #=> count_vowels("colour") == 3: true #=> count_vowels("cecilia") == 4: true
我认为使用HashTable数据结构将是解决这个特定问题的好方法。 特别是如果你需要分别输出每个元音的数量。
这是我使用的代码：
def vowels(string) found_vowels = Hash.new(0) string.split("").each do |char| case char.downcase when 'a' found_vowels['a']+=1 when 'e' found_vowels['e']+=1 when 'i' found_vowels['i']+=1 when 'o' found_vowels['o']+=1 when 'u' found_vowels['u']+=1 end end found_vowels end p vowels("aeiou")
甚至这个（优雅但不一定高性能）：
def elegant_vowels(string) found_vowels = Hash.new(0) string.split("").each do |char| case char.downcase when ->(n) { ['a','e','i','o','u'].include?(n) } found_vowels[char]+=1 end end found_vowels end p elegant_vowels("aeiou")
哪个会输出：
{“a”=> 1，“e”=> 1，“i”=> 1，“o”=> 1，“u”=> 1}
因此，您不必将字符串转换为数组并担心区分大小写：
def getVowelCount(string) string.downcase.count 'aeiou' end