我必须创建一个程序，接受用户输入的两个单词，并从这两个提取唯一的字母.
两个词是"外推"和"中间".
go 掉常见字母后，无论输入单词的顺序如何，结果都应该是"外推".在外推法中，虽然有两个‘a’，但只go 掉了一个..
我写的代码直到现在都是在
puts "Enter the first_word : "
word_1 = gets.chomp.downcase.split('', 0)
puts "Enter second word name: "
word_2 = gets.chomp.downcase.split('', 0)
p word_1
p word_2
differences = word_1- word_2
p differences
#input1 = among
#input2 = extrapolation
# output => ["m","g"]
代码忽略第二个单词中不在第一个单词中的字母.
有没有内置的字符串方法来处理这个问题？
提前感谢你的帮助..
假设给我们两个词.
w1 = "extrapolation"
w2 = "among"
Step 1: Convert each of these strings to an array of characters个
a1 = w1.chars
#=> ["e", "x", "t", "r", "a", "p", "o", "l", "a", "t", "i", "o", "n"]
a2 = w2.chars
#=> ["a", "m", "o", "n", "g"]
Step 2: Create a hash for each array that gives the index of the first instance of each unique character in the array个
def first_char_pos(a)
a.each_with_index.with_object({}) do |(c,i),h|
h[c] = i unless h.key?(c)
end
end
h1 = first_char_pos(a1)
#=> {"e"=>0, "x"=>1, "t"=>2, "r"=>3, "a"=>4,
# "p"=>5, "o"=>6, "l"=>7, "i"=>10, "n"=>12}
h2 = first_char_pos(a2)
#=>{"a"=>0, "m"=>1, "o"=>2, "n"=>3, "g"=>4}
Step 3: Identify the letters to be removed from each word个
letters_to_remove = a1 & a2
#=> ["a", "o", "n"]
Step 4: Identify the indices of the letters in each word to keep个
def indices_to_keep(a, h, letters_to_remove)
a.size.times.to_a - h.values_at(*letters_to_remove)
end
idx1 = indices_to_keep(a1, h1, letters_to_remove)
#=>[0, 1, 2, 3, 5, 7, 8, 9, 10, 11]
idx2 = indices_to_keep(a2, h2, letters_to_remove)
#=> [1, 4]
Step 5: Identify, in order, the letters in each word to keep个
s1 = a1.values_at(*idx1)
#=> ["e", "x", "t", "r", "p", "l", "a", "t", "i", "o"]
s2 = a2.values_at(*idx2)
#=> ["m", "g"]
Step 6: Form the desired string个
s1.join + s2.join
#=> "extrplatiomg"
这与所需的字符串不同，后者为"extrpolatimg".原因是我删除了要删除的每个字母的第一个实例，而所需的字符串是通过删除每个字符串的第一个"a"和最后一个"o"来获得的.要么是OP在给出想要的结果时出错，要么是我不理解确定要删除字母的哪个实例的规则.
我没有提到可能出现大写字母的情况，因为我不知道在这种情况下适用的规则.