Codewars' Kata

2017年3月25日 · 阅读需 7 分钟

Chengyu HAN

Open Source Contributor

看到 Codewars oj类似物，不过可以看到详细的报错，并在AC一次后看到别人的答案

开坑记录一下有趣的题, 顺便采用新结构 [堆栈类似物]

我的解题集

1ku

2ku

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4ku

Roman Numerals Decoder

根据以前找的js，改写一下带验证的函数

def Roman_Num_Decoder(s)
  validator = /^M*(?:D?C{0,3}|C[MD])(?:L?X{0,3}|X[CL])(?:V?I{0,3}|I[XV])$/
  token = /[MDLV]|C[MD]?|X[CL]?|I[XV]?/
	key = { M:1000,
          CM:900, D:500, CD:400, C:100,
          XC:90,  L:50,  XL:40,  X:10,
          IX:9,   V:5,   IV:4,   I:1}

  
  s.upcase.split(' ').map do |rnum|
    if validator.match(rnum)
      rnum.scan(token).
      reduce(0) {|sum, rn| sum+=key[rn.to_sym]}
    else
      puts "Not valid Roman Numbers!\n"
      return nil
    end
  end
end

s = 'CI XCVII CXV CXXI XCIX CXVI CII CXXIII CXVI CIV CV CXV XCV CV CXV XCV CX CXI CXVI XCV CXVI CIV CI XCV CII CVIII XCVII CIII CXXV XXXII CV XXXII CVII CX CXI CXIX XXXII CXVI CIV CV CXV XXXII CII CVIII XCVII CIII XXXII CV CXV XXXII CXIX CI CV CXIV C XXXII CV XXXII CVI CXVII CXV CXVI XXXII XCIX XCVII CX XXXIX CXVI XXXII CXII CXVII CXVI XXXII CIX CXXI XXXII CII CV CX CIII CI CXIV XXXII CXI CX XXXII CXIX CIV CXXI'
p Roman_Num_Decoder(s)

题目要求的无检验

def Roman_Num_Decoder(s)
  token = /[MDLV]|C[MD]?|X[CL]?|I[XV]?/
	key = { M:1000,
          CM:900, D:500, CD:400, C:100,
          XC:90,  L:50,  XL:40,  X:10,
          IX:9,   V:5,   IV:4,   I:1}
  
  s.
    upcase.
    scan(token).
    reduce(0) {|sum, rn| sum+=key[rn.to_sym]}
end

s = 'CI XCVII CXV CXXI XCIX CXVI CII CXXIII CXVI CIV CV CXV XCV CV CXV XCV CX CXI CXVI XCV CXVI CIV CI XCV CII CVIII XCVII CIII CXXV XXXII CV XXXII CVII CX CXI CXIX XXXII CXVI CIV CV CXV XXXII CII CVIII XCVII CIII XXXII CV CXV XXXII CXIX CI CV CXIV C XXXII CV XXXII CVI CXVII CXV CXVI XXXII XCIX XCVII CX XXXIX CXVI XXXII CXII CXVII CXVI XXXII CIX CXXI XXXII CII CV CX CIII CI CXIV XXXII CXI CX XXXII CXIX CIV CXXI'

s.split(' ').each do |rn|
  print Roman_Num_Decoder(rn), ', '
end
print "\n"

5ku

"Simple Fun #299: Look And Say And Sum" Translation

Solution

module Haskell.Codewars.LookAndSayAndSum where
import Data.List (group)
import Data.Char (digitToInt)

lookAndSay' :: Integer -> Integer
lookAndSay' n = read (concatMap count (group (show n)))
  where 
    count run = show (length run) ++ take 1 run

toDigits = map (fromIntegral . digitToInt) . show

lookAndSaySum :: Int -> Int
lookAndSaySum n = foldr (+) 0 (toDigits (iterate lookAndSay' 1 !! (n - 1)) )

Test Case

import Test.Hspec
import Haskell.Codewars.LookAndSayAndSum (lookAndSaySum)
import System.Random

sol :: Int -> Int
sol n = [0,1,2,3,5,8,10,13,16,23,32,44,56,76,102,132,174,227,296,383,505,679,892,1151,1516,1988,2602,3400,4410,5759,7519,9809,12810,16710,21758,28356,36955,48189,62805,81803,106647,139088,181301,236453,308150,401689,523719,682571,889807,1159977,1511915,1970964,2569494,3349648,4366359,5691884] !! n

randint :: IO Int
randint = randomRIO (1,55)

main :: IO ()
main = hspec $ do
  describe "Unit tests" $ do
    it "Basic tests" $ do
      lookAndSaySum 1 `shouldBe` 1
      lookAndSaySum 2 `shouldBe` 2
      lookAndSaySum 3 `shouldBe` 3
      lookAndSaySum 4 `shouldBe` 5
      lookAndSaySum 5 `shouldBe` 8
    it "Radnom tests" $ do
      rint <- sequence (replicate 5 randint) 
      map lookAndSaySum rint `shouldBe` map sol rint

First Variation on Caesar Cipher

5ku的题果然不是那么简单的先写一个凯撒函数

def movingShift(s, shift)
  s.gsub!(/[A-Z]/) do |c| 
    tmp = c.ord.+(shift%27).%('Z'.ord+1)
    tmp += 'A'.ord unless tmp >= c.ord
    tmp.chr
  end
  s.gsub!(/[a-z]/) do |c| 
    tmp = c.ord.+(shift%27).%('z'.ord+1)
    tmp += 'a'.ord unless tmp >= c.ord
    tmp.chr
  end
end

def demovingShift(arr, shift)
  movingShift(arr.join, 26-shift)
end


p u = "I should have known that you would have a perfect answer for me!!!"
p v = movingShift(u, 1)
p demovingShift(v, 1)

Simplify

def movingShift(s, shift)
  t = s.chars.join    # deep clone 保证输入字符串s不变
  t.gsub!(/\w/) do |c| 
    isUC = c.ord < 91
    tmp = c.ord.+(shift%26).%((isUC ? 'Z' : 'z').ord+1)
    tmp += (isUC ? 'A' : 'a').ord unless tmp >= c.ord
    tmp.chr
  end
end

按题意修改程序，并格式化输出

def movingShift(s, shift)
  s.chars.map.with_index do |c, i|
    if c =~ /\w/
      isUC = c.ord < 91
      tmp = c.ord.+((i+shift)%26).%((isUC ? 'Z' : 'z').ord+1)
      tmp += (isUC ? 'A' : 'a').ord unless tmp >= c.ord
      tmp.chr
    else
      c
    end
  end.each_slice(s.length/5+1).to_a.map{|a| a.join}
end

def demovingShift(arr, shift)
  arr.join.chars.map.with_index do |c, i|
    if c =~ /\w/
      movingShift(c, 26-shift-i)
    else
      c
    end
  end.join
end

解决数字干扰问题，并添加奇怪的输出格式支持

def movingShift(s, shift)
  len = s.length
  @obj = nil
  s.chars.map.with_index do |c, i|
    if c =~ /[a-z]/i
      isUC = c.ord < 91
      tmp = c.ord.+((i+shift)%26).%((isUC ? 'Z' : 'z').ord+1)
      tmp += (isUC ? 'A' : 'a').ord unless tmp >= c.ord
      tmp.chr
    else
      c
    end
  end.
  each_slice(len%5 == 0 ? len/5: len/5+1).
  to_a.map{|a| a.join}.
  tap{|obj| @obj = obj}.            # 从链式调用中 ‘抽样’
  +(@obj.length == 5 ? [] : [''])   # 保证返回的数组有5个元素
end

def demovingShift(arr, shift)
  arr.join.chars.map.with_index do |c, i|
    if c =~ /[a-z]/i
      movingShift(c, 26-shift-i)
    else
      c
    end
  end.join
end

6ku

Dubstep

def song_decoder(song)
  song.gsub(/(WUB)+/, ' ').strip
end

GET 新函数strip,去除字符串前后空白字符.

strip → new_str click to toggle source
Returns a copy of str with leading and trailing whitespace removed.

Take a Number And Sum Its Digits Raised To The Consecutive Powers And ....¡Eureka!!

GET：

知道了 enumerate <- 不记得枚举，最后写了个循环

My Solution - Reduce ver

def sum_dig_pow(a, b): 
    return filter(
        lambda g:
            reduce(
                lambda y,x: y+int(str(g)[x])**(x+1), range(0, len(str(g))), 0
            ) == g
        , range(a, b+1))

enumerate ver

def sum_dig_pow(a, b): 
    return filter(
        lambda g:
            sum(
                int(d) ** (i+1) 
                for i, d in enumerate(str(a))
            ) == a
        , range(a, b+1))

7ku

a.map{|arr| arr.map{|a| avg = a.reduce(:+)./3; [avg, avg, avg] }}

Ones and Zeros

GET:

join的玩法

Reduce ver

from functools import reduce
def binary_array_to_number(arr):
    return int(reduce(lambda y,x:y+str(x), arr, ''), 2)

join ver

def binary_array_to_number(arr):
    return int("".join(map(str, arr)), 2)

Regex validate PIN code

GET:

正则表达式姿势 ^[0-9]{4}$+^[0-9]{6}$==^(\d{4}|\d{6})$

Get the Middle Character

GET:

复习了Python 的整除

Sum of two lowest positive integers

GET：

sort&sorted 的用法

a = [5,2,1,9,6]

>>> sorted(a) #将a从小到大排序,不影响a本身结构
[1, 2, 5, 6, 9]

>>> sorted(a,reverse = True) #将a从大到小排序,不影响a本身结构
[9, 6, 5, 2, 1]

>>> a.sort() #将a从小到大排序,影响a本身结构
>>> a
[1, 2, 5, 6, 9]

>>> a.sort(reverse = True) #将a从大到小排序,影响a本身结构
>>> a
[9, 6, 5, 2, 1]

8ku

Basic Mathematical Operations

Get:

switch-case替代方案

def basic_op(operator, value1, value2):
    ops = {'+': lambda a, b: a + b, 
           '-': lambda a, b: a - b,
           '*': lambda a, b: a * b,
           '/': lambda a, b: a / b}
    return ops[operator](value1, value2)

Grasshopper - Summation

GET:

reduce&lambda表达式 初体验

Remove First and Last Character

GET:

切片技巧

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