如何比较两个字符串中的字母Ruby

Question

我是 ruby​​ 的新手，我正在创建一个刽子手游戏。到目前为止，我的代码将单词与正确的单词进行了比较。但我希望它比较字母。所以基本上，如果秘密词是glue，用户输入G它会不正确，但如果用户输入glue它会是正确的。我需要它像刽子手一样逐字比较。

对此有点麻烦。我在下面附上了我的代码。

secret_word = []
puts "enter a word to be guessed"
secret_word = gets.chomp



guess_letters = []
guess = ""
guess_limit = 3
guess_count = 0
out_of_guesses = false

while guess != secret_word and !out_of_guesses
    if guess_count < guess_limit
    puts "enter your guess: "
    guess = gets.chomp()
    guess_letters << guess
    guess_count +=1
    puts "you have used these letters thus far #{guess_letters.join(", ")}"
else
    out_of_guesses = true
end
end
if out_of_guesses
    puts "you Lose, the word was #{secret_word}"
else
    puts "you win"
end

Answer 1

我不确定您使用的是哪个刽子手规则，但这是一个粗略的草案，允许 3 次失败的尝试并使用小写字符

def guess_word(word, tries)

    if tries < 1
        puts "You are hanged!"
    elsif word.empty?
        puts "You guessed it! You are saved from the gallows!"
    else
        print "Enter character: "
        c = STDIN.getc.downcase
        STDIN.getc # get rid of newline
        if word.index(c).nil?
            puts "Ooops, #{c} was wrong!"
            guess_word(word, tries - 1)
        else
            puts "#{c} was correct!"
            guess_word(word.sub(/["#{c}"]/, ''), tries)
        end
    end

end

if __FILE__ == $0

    TRIES = 3

    print "Enter word to guess: "
    word = gets.chomp

    guess_word(word.downcase, 3)    

end

这是未经测试的..

Answer 2

游戏规则hangman在其Wiki 给出。我假设当绞刑架上的人的所有七个部分（头部、颈部、左臂、身体、右臂、左腿、右腿）都被画出时，试图猜测这个词的玩家会输。

辅助方法

画出被吊死的人

首先创建一个可以用来绘制部分或全部刽子手的散列：

MAN = [" O\n", " |\n", "\\", "|", "/\n", " |\n/", " \\"].
  map.each_with_object([""]) { |s,arr| arr << (arr.last + s) }.
      each.with_index.with_object({}) { |(s,i),h| h[i] = s }

关键是猜错的次数。例如：

puts MAN[2]
 O
 |
puts MAN[6]
 O
 |
\|/
 |
/

跟踪单词的字母位置

接下来创建一个散列，其键是秘密单词的唯一字母，其值是单词中键位置的索引数组。

def construct_unknown(word)
  word.each_char.with_index.with_object({}) { |(c,i),h| (h[c] ||= []) << i }
end

例如，

unknown = construct_unknown("beetle")
  #=> {"b"=>[0], "e"=>[1, 2, 5], "t"=>[3], "l"=>[4]}

我们还将为位置已知的字母创建一个空哈希：

known = {}

将猜测的字母从哈希 unknown 移动到哈希 known

如果猜测的字母是unknown 的键，则该键和值将移动到known。

def move_unknown_to_known(letter, unknown, known)
  known.update(letter=>unknown[letter])
  unknown.delete(letter)
end

例如（对于上面的unknown 和known），

move_unknown_to_known("e", unknown, known)
unknown #=> {"b"=>[0], "t"=>[3], "l"=>[4]} 
known   #=> {"e"=>[1, 2, 5]} 

看看猜的人是赢了还是输了

我们在猜出一个字母后确定玩家何时赢或输，或继续：

def win?(word_size, known)
  known.values.flatten.sum == word_size
end

def lose?(wrong_guess_count)
  wrong_guess_count == HANGMAN.size
end

例如，

win?(word.size, known)
  #=> false

lose?(6) #=> false
lose?(7) #=> true

显示已知字母

def display_known(word_size, known)
  known.each_with_object('_' * word_size) { |(k,a),s| a.each { |i| s[i] = k } }
end

例如（召回word #=> "beetle"），

puts display_known(word.size, known)
_ee__e

主要方法

我们现在可以编写 main 方法了。

def hangman
  puts "Player 2, please avert your eyes for a moment."
  print "Player 1: enter a secret word with at least two letters: "
  word = gets.chomp.downcase
  unknown = construct_unknown(word)
  known = {}
  wrong_guess_count = 0
  loop do
    puts display_known(word.size, known)
    puts MAN[wrong_guess_count] if wrong_guess_count > 0
    if win?(word.size, known)
      puts "You win! You win! Congratulations!"
      break  
    end
    if lose?(wrong_guess_count)
      puts "Sorry, but you've run out of guesses"
      break
    end 
    print "Player 2: enter a letter or your guess of the word: "
    guess = gets.chomp.downcase
    if guess.size > 1
      if guess == word
        puts word
        puts "You win! You win! Congratulations!"
        break
      else
        puts "Sorry, that's not the word"
        wrong_guess_count += 1
      end
    elsif unknown.key?(guess)
      nbr = unknown[guess].size
      puts nbr == 1 ? "There is 1 #{guess}" : "There are #{nbr} #{guess}'s"
      move_unknown_to_known(guess, unknown, known)                   
    else
      puts "Sorry, the word contains no #{guess}'s"
      wrong_guess_count += 1
    end
  end
end  

示例

在向两位选手和观众解释了规则后，嘉宾主持人最后说：“别忘了，在猜一个字母或单词时，必须将其表达为一个问题……片刻。 ..持有那个...有人告诉我没有必要将其作为一个问题提出来”。

假设单词是beetle，字母猜测是't'、'i'、'a'、'l'、'r'、's'、't'、'u'、@986 , 'beetle'.

hangman

Player 2, please avert your eyes for a moment.

Player 1: enter a secret word with at least two letters: beetle
______

Player 2: enter a letter or your guess of the word: t
There is 1 t
___t__

Player 2: enter a letter or your guess of the word: i
Sorry, the word contains no i's
___t__
 O

Player 2: enter a letter or your guess of the word: a
Sorry, the word contains no a's
___t__
 O
 |

Player 2: enter a letter or your guess of the word: l
There is 1 l
___tl_
 O
 |

Player 2: enter a letter or your guess of the word: r
Sorry, the word contains no r's
___tl_
 O
 |
\

Player 2: enter a letter or your guess of the word: s
Sorry, the word contains no s's
___tl_
 O
 |
\|

Player 2: enter a letter or your guess of the word: t
Sorry, the word contains no t's
___tl_
 O
 |
\|/

Player 2: enter a letter or your guess of the word: u
Sorry, the word contains no u's
___tl_
 O
 |
\|/
 |
/

Player 2: enter a letter or your guess of the word: e
There are 3 e's
_eetle
 O
 |
\|/
 |
/

Player 2: enter a letter or your guess of the word: beetle
beetle
You win! You win! Congratulations!