实现一个方法，该方法采用 2 个参数并返回一个哈希值，其中包含子字符串在字符串中出现的次数

Question

我目前正在通过 odin 项目课程学习 Ruby。这是项目描述：

实现一个方法

#substrings

，将单词作为第一个参数，然后将有效子字符串数组（您的字典）作为第二个参数。它应该返回一个哈希值，列出在原始字符串中找到的每个子字符串（不区分大小写）以及找到的次数。

dictionary = ["below","down","go","going","horn","how","howdy","it","i","low","own","part","partner","sit"]
# => ["below","down","go","going","horn","how","howdy","it","i","low","own","part","partner","sit"]

substrings("below", dictionary)
# => { "below" => 1, "low" => 1 }

substrings("Howdy partner, sit down! How's it going?", dictionary)
#  => { "down" => 1, "go" => 1, "going" => 1, "how" => 2, "howdy" => 1, "it" => 2, "i" => 3, "own" => 1, "part" => 1, "partner" => 1, "sit" => 1 }

我首先将字符串转为数组，这样每个单词都是一个数组元素。我循环遍历字典，然后循环遍历字符串数组来比较两个单词，看看它是否包含字典中的子字符串。我使用

string[substring]

检查了子字符串。如果没有返回 nil （意味着字符串包含字典中的子字符串，我会将其推入我的哈希中，创建一个新的键值对。

def substrings(string, dictionary)
  result = Hash.new
  string_array = string.split" "

  dictionary.each do |dic_word|
    string_array.each do |string_word|
      string_word = string_word.downcase
      if string_word[dic_word] != nil
        if result.key?(dic_word)
          result[dic_word] += 1
        else
          result[dic_word] = 1
        end
      end
    end
  end
  result
end

该方法运行良好，我的问题是关于时间复杂度和 Ruby 中的最佳实践。有没有办法提高效率？或者也许有一种方法可以让它看起来更干净，使用其他方法可以更好地解决这个问题。

有关我的代码可读性、变量名称等的任何其他建设性反馈将不胜感激。

Answer 1

我发现 ruby 有一个 #scan 方法，可以提高性能并且不需要新数组

  def substrings(string, dictionary)
  result = Hash.new
  string = string.downcase
  dictionary.each do |word|
    result[word] = string.scan(word).length unless string.scan(word).length == 0
  end

  result
end