如何找到重复字母最多的单词 [英] How to find word with the greatest number of repeated letters

问题描述

我的目标是找到给定字符串中重复字母最多的单词.例如，"aabcc ddeeteefef iijjfff" 将返回 "ddeeteefef" 因为 "e" 在这个单词中重复了五次，并且超过了所有其他重复字符.

My goal is to find the word with greatest number of repeated letters in a given string. For example, "aabcc ddeeteefef iijjfff" would return "ddeeteefef" because "e" is repeated five times in this word and that is more than all other repeating characters.

到目前为止，这是我得到的，但它有很多问题并且不完整:

So far this is what I got, but it has many problems and is not complete:

def LetterCountI(str)
  s = str.split(" ")
  i = 0
  result = []
  t = s[i].scan(/((.)\2+)/).map(&:max) 
  u = t.max { |a, b| a.length <=> b.length }
  return u.split(//).count 
end

我的代码只能找到连续的模式；如果模式被打断(例如使用 "aabaaa"，它会计数 3 次而不是 5 次).

The code I have only finds consecutive patterns; if the pattern is interrupted (such as with "aabaaa", it counts a three times instead of five).

推荐答案

我会这样做:

s = "aabcc ddeeteefef iijjfff" 
# intermediate calculation that's happening in the final code
s.split(" ").map { |w| w.chars.max_by { |e| w.count(e) } }
# => ["a", "e", "f"] # getting the max count character from each word
s.split(" ").map { |w| w.count(w.chars.max_by { |e| w.count(e) }) }
# => [2, 5, 3] # getting the max count character's count from each word
# final code
s.split(" ").max_by { |w| w.count(w.chars.max_by { |e| w.count(e) }) }
# => "ddeeteefef"

更新

each_with_object 比 group_by 方法给出更好的结果.

each_with_object gives better result than group_by method.

require 'benchmark'

s = "aabcc ddeeteefef iijjfff" 

def phrogz(s)
   s.scan(/\w+/).max_by{ |word| word.chars.group_by(&:to_s).values.map(&:size).max }
end

def arup_v1(s)
    max_string = s.split.max_by do |w| 
       h = w.chars.each_with_object(Hash.new(0)) do |e,hsh|
         hsh[e] += 1
       end
       h.values.max
    end
end

def arup_v2(s)
   s.split.max_by { |w| w.count(w.chars.max_by { |e| w.count(e) }) }
end

n = 100_000
Benchmark.bm do |x|
  x.report("Phrogz:") { n.times {|i| phrogz s } }
  x.report("arup_v2:"){ n.times {|i| arup_v2 s } }
  x.report("arup_v1:"){ n.times {|i| arup_v1 s } }
end

输出

            user     system      total        real
Phrogz:   1.981000   0.000000   1.981000 (  1.979198)
arup_v2:  0.874000   0.000000   0.874000 (  0.878088)
arup_v1:  1.684000   0.000000   1.684000 (  1.685168)

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