Question

我在Ruby中找到了后缀数组的实现并稍微改了一下。这就是我所拥有的：

class SuffixArray
    def initialize(str)
        @string = str
        @suffix_array = []
        (0...str.length).each do |i|
            substring = @string[i...str.length]
            @suffix_array << {:suffix=>substring, :index => i}
        end

        @sorted_suffix_array = @suffix_array.sort {|x,y| x[:suffix] <=> y[:suffix]}
    end

    def print_sorted
      @sorted_suffix_array.each {|item| puts "#{item[:index]}=>#{item[:suffix]}"}
      puts "total=>#{@sorted_suffix_array.size()}"
    end

    def print_unsorted
      @suffix_array.each {|item| puts "#{item[:index]}=>#{item[:suffix]}"}
      puts "total=>#{@suffix_array.size()}"
    end

    def find_substring(substring)
        low = 0
        high = @sorted_suffix_array.length
        while(low <= high) do
            mid = (low + high) / 2
            comparison = @sorted_suffix_array[mid][:suffix]#[0..substring.length]
      if comparison > substring
        high = mid - 1
      elsif comparison < substring
        low = mid + 1
      else 
        return @sorted_suffix_array[mid][:index]
      end
        end
    end

end

它运作良好，但找不到我想要的所有子串。例如

a = SuffixArray.new("there is a man who likes dogs")
puts a.find_substring("man") #won't be found
puts a.find_substring("who likes dogs") #will be found
puts a.find_substring("o likes dogs") #will be found

如何更改算法以使其找到我想要的所有子字符串？

Answer 1

您的代码几乎是正确的。我做了一些小修改，但它确实有效。

def find_substring(substring)
  low = 0
  high = @sorted_suffix_array.length-1
  while(low <= high) do
    mid = (low + high) / 2
    comparison = @sorted_suffix_array[mid][:suffix][0...substring.length]
    if comparison > substring
      high = mid - 1
    elsif comparison < substring
      low = mid + 1
    else 
      return @sorted_suffix_array[mid][:index]
    end
  end
end

Answer 2

对于其他人;引用，这是没有将子字符串保存在哈希

中的引用

要点：https://gist.github.com/bluetwin/5268722

class SuffixArray

  attr_reader :suf, :string

  def initialize(string)
    @string = string
    @suf = (0..string.size-1).sort_by{|i|@string[i..-1]}
  end

  def substring(idx)
    @string[@suf[idx]..@string.size-1]
  end

  def bsearch(str)
    low = 0
    high = @suf.length-1
    found = nil
    while(low <= high) do
      mid = (low + high) / 2
      comp = substring(mid)
      if comp > str
        high = mid - 1
      elsif comp < str
        low = mid + 1
      else
       found = comp
       low = high + 1
      end
    end
    found
  end

end

后缀数组并搜索字符串中的子字符串

2 个答案: