如何以一切可能的方式將字符串拆分為長度最多為3的連續子串？

Question

我試圖在長度1和10之間取一個字符串，並輸出所有可能的方法將字符串分解為大小為1,2或3的連續子字符串。例如：

Input: 123456

將整數切成單個字符，然后繼續查找組合。 代碼將返回以下所有數組。

    [1, 2, 3, 4, 5, 6]  
    [12, 3, 4, 5, 6]  
    [1, 23, 4, 5, 6]  
    [1, 2, 34, 5, 6]  
    [1, 2, 3, 45, 6]  
    [1, 2, 3, 4, 56]  
    [12, 34, 5, 6]  
    [12, 3, 45, 6]  
    [12, 3, 4, 56]  
    [1, 23, 45, 6]  
    [1, 2, 34, 56]  
    [1, 23, 4, 56]  
    [12, 34, 56]  
    [123, 4, 5, 6]  
    [1, 234, 5, 6]  
    [1, 2, 345, 6]  
    [1, 2, 3, 456]  
    [123, 456]  
    [1, 23, 456]  
    [1, 234, 56]  
    [12, 345, 6]  
    [12, 3, 456]  
    [123, 4, 56]  
    [123, 45, 6]

我想用紅寶石做這件事。 謝謝！

Answer 1

這是一個有效的功能。 可能不是最佳的，因為我沒有花太多時間。

str = "1234567890"

def f(s, n)
    return [[]] if s.empty?

    (1..[n, s.length].min).map{|c| f(s[c..-1], n).map{|a| [s[0, c]] + a}}.inject(&:+)
end

puts f(str, 3).collect{|l| l * "\t"}

編輯：使它縮短一點，現在作為第二個參數傳遞長度以實現靈活性。

Answer 2

我花了很長時間來弄明白這一點，這是我第一次遇到的更難的問題。 但最終我遇到了這個解決方案：

def combinations(s)
  c = (s.length > 3) ? [] : [[s]]
  max = [4, s.length].min
  (1...max).inject(c) do |c, i|
    combinations(s[i..-1]).inject(c) do |c, tail|
      c.push([s[0...i]] + tail)
    end
  end
end

combinations("12345").each { |c| p c }

生產：

["1", "2", "345"]
["1", "2", "3", "45"]
["1", "2", "3", "4", "5"]
["1", "2", "34", "5"]
["1", "23", "45"]
["1", "23", "4", "5"]
["1", "234", "5"]
["12", "345"]
["12", "3", "45"]
["12", "3", "4", "5"]
["12", "34", "5"]
["123", "45"]
["123", "4", "5"]

Answer 3

這是另一個：

class String
  def splitup(prefix=nil)
    parts = []
    if size <= 3
      parts << [prefix,self].compact * ","
    end
    (1..([size,3].min)).each do |p|
      next if p >= size
      parts << slice(p..-1).splitup([prefix,slice(0,p)].compact * ",")
    end
    parts
  end

  def report
    flat = splitup.flatten.sort_by {|x| [-x.size,x]}
    puts
    puts "#{flat.size} permutations of #{self}"
    puts flat
  end
end

然后

>> "123456".report

24 permutations of 123456
1,2,3,4,5,6
1,2,3,4,56
1,2,3,45,6
1,2,34,5,6
1,23,4,5,6
12,3,4,5,6
1,2,3,456
1,2,34,56
1,2,345,6
1,23,4,56
1,23,45,6
1,234,5,6
12,3,4,56
12,3,45,6
12,34,5,6
123,4,5,6
1,23,456
1,234,56
12,3,456
12,34,56
12,345,6
123,4,56
123,45,6
123,456

Answer 4

def f(s, mx)
  (1..[s.size, mx].min).each_with_object([]) do |n,a|
     if n < s.size
       sf = s[0, n]
       f(s[n..-1], mx).each { |c| a << [sf, *c] }
     else
       a << [s]
     end
  end
end

f("12345", 3)
  #=> [["1", "2", "3", "4", "5"], ["1", "2", "3", "45"],
  #    ["1", "2", "34", "5"], ["1", "2", "345"], ["1", "23", "4", "5"],
  #    ["1", "23", "45"], ["1", "234", "5"], ["12", "3", "4", "5"],
  #    ["12", "3", "45"], ["12", "34", "5"], ["12", "345"],
  #    ["123", "4", "5"], ["123", "45"]] => [["1", "2", "3", "4"],
  #    ["1", "2", "34"], ["1", "23", "4"], ["1", "234"],
  #    ["12", "3", "4"], ["12", "34"], ["123", "4"]]