Ruby:字符串重建算法，只能部分工作

Question

我正在Ruby中研究一种字符串重建算法（动态编程示例中的经典算法，将无空格文本转换为标准空格文本）下面的代码是纯ruby，您可以立即复制粘贴并开始测试，它80%的时间都在工作，而且字典越大，它就越容易崩溃。我用超过8万个单词的词典测试过它，但它的效果不太好，大约70%的时间。

如果有一种方法可以让它100%工作，如果这个词出现在字典里，请给我看看

这是代码：（它间隔很好，应该很可读）

# Partially working string reconstruction algo in pure Ruby
# the dictionary
def dict(someWord)
  myArray = [" ", "best", "domain", "my", "successes", "image", "resizer", "high", "tech", "crime", "unit", "name", "edge", "times", "find", "a", "bargain", "free", "spirited", "style", "i", "command", "go", "direct", "to", "harness", "the", "force"]
  return !!(myArray.index(someWord))
end

# inspired by http://cseweb.ucsd.edu/classes/wi12/cse202-a/lecture6-final.pdf
## Please uncomment the one you wanna use
#
# (all the words used are present in the dictionary above)
#
# working sentences
  x = ' ' + "harnesstheforce"
# x = ' ' + "hightechcrimeunit"
#
# non working sentences
# x = ' ' + "findabargain"
# x = ' ' + "icommand"
puts "Trying to reconstruct #{x}"
# useful variables we're going to use in our algo
n = x.length
k = Array.new(n)
s = Array.new(n)
breakpoints = Hash.new
validBreakpoints = Hash.new
begin
  # let's fill k
  for i in 0..n-1
    k[i] = i
  end
  # the core algo starts here
  s[0] = true
  for k in 1..n-1
    s[k] = false
    for j in 1..k
      if s[j-1] && dict(x[j..k])
        s[k] = true
        # using a hash is just a trick to not have duplicates
        breakpoints.store(k, true)
      end
    end
  end
  # debug
  puts "breakpoints: #{breakpoints.inspect} for #{x}"
  # let's create a valid break point vector
  i=1
  while i <= n-1 do
    # we choose the longest valid word
    breakpoints.keys.sort.each do |k|
      if i >= k
        next
      end
      # debug: when the algo breaks, it does so here and goes into an infinite loop
      #puts "x[#{i}..#{k}]: #{x[i..k]}"
      if dict(x[i..k])
        validBreakpoints[i] = k
      end
    end
    if validBreakpoints[i]
      i = validBreakpoints[i] + 1
    end
  end
  # debug
  puts "validBreakpoints: #{validBreakpoints.inspect} for #{x}"
  # we insert the spaces at the places defined by the valid breakpoints
  x = x.strip
  i = 0
  validBreakpoints.each_key do |key|
    validBreakpoints[key] = validBreakpoints[key] + i
    i += 1
  end
  validBreakpoints.each_value do |value|
    x.insert(value, ' ')
  end
  puts "Debug: x: #{x}"

  # we capture ctrl-c
rescue SignalException
  abort
# end of rescue
end

Answer 1

请注意，对于包含单字符单词的字符串，您的算法会失败。这是一个错误。您忽略了这些单词后面的断点，因此您最终得到的单词（"abargain"）不包含在字典中。

更改

   if i >= k
     next
   end

至

   if i > k
     next
   end

或更多类似Ruby的

   next if i > k

还需要注意的是，每当字符串包含非单词的东西时，就会陷入一个无休止的循环：

if validBreakpoints[i]        # will be false 
  i = validBreakpoints[i] + 1 # i not incremented, so start over at the same position
end

你最好将此视为错误

return '<no parse>' unless validBreakpoints[i] # or throw if you are not in a function
i = validBreakpoints[i] + 1

"inotifier"的问题是您的算法的不足。总是选择最长的单词是不好的。在这种情况下，检测到的第一个"有效"断点位于"in"之后，这会留下非单词"otifier"。