1、Simple Pig Latin
Move the first letter of each word to the end of it, then add "ay" to the end of the word. Leave punctuation marks untouched.
Examples
pig_it('Pig latin is cool') # igPay atinlay siay oolcay
pig_it('Hello world !') # elloHay orldway !解答：
def pig_it(text):
strList = text.split()
for i in xrange(len(strList)):
if strList[i].isalpha():
strList[i] = strList[i][1:len(strList[i])] + strList[i][0] + 'ay'
return ' '.join(strList)2、 Calculate Fibonacci return count of digit occurrences
A farmer raises rabbits. Each rabbit gives birth to one rabbit when it turns 2 months old, and then to one rabbit each month. Rabbits never die, and we ignore hares. How many rabbits will the farmer have in the n-th month, if he starts with one newborn rabbit?
Fibonacci numbers are generated by setting F0=0, F1=1, and then using the formula
Fn = Fn-1 + Fn-2
Your task is to efficiently calculate the nth element in the Fibonacci sequence and then count the occurrence of each digit in the number returning a list of integer pairs sorted in descending order.
10 ≤ N ≤ 100000
f(10)=55 returns
a = [(2, 5)]
two occurances of the digit 5
f(10)=55
^^
||
f(10000) returns
a = [(254, 3),
(228, 2),
(217, 6),
(217, 0),
(202, 5),
(199, 1),
(198, 7),
(197, 8),
(194, 4),
(184, 9)]
If two integers have the same count sort them in descending order.
Your algorithm must be efficient.
解答：
def calc(n):
if n == 0:
return (0, 1)
elif n == 1:
return (1, 1)
else:
a, b = calc(n // 2)
p = a * (2 * b - a)
q = b * b + a * a
return (p, q) if n % 2 == 0 else (q, p + q)
# 一种计算斐波那契数列第n个值的高效算法，可计算超大数
def fib(n):
if n >= 0:
return calc(n)[0]
else:
return -calc(-n)[0] if n % 2 ==0 else calc(-n)[0]
# 排序规则
def takeFirst(elem):
return elem[0]
def fib_digits(n):
res = []
for i in xrange(10):
if str(fib(n)).count(str(i)) > 0:
res.append((str(fib(n)).count(str(i)), i))
res.sort(key = takeFirst, reverse = True)
if len(res) >= 2:
for i in xrange(len(res)-1):
if res[i][0] == res[i+1][0]:
res[i],res[i+1] = res[i+1],res[i]
return res3、Encrypt this!
You want to create secret messages which can be deciphered by the Decipher this! kata. Here are the conditions:
- Your message is a string containing space separated words.
- You need to encrypt each word in the message using the following rules:
- The first letter needs to be converted to its ASCII code.
- The second letter needs to be switched with the last letter
- Keepin' it simple: There are no special characters in input.
Examples:
encrypt_this("Hello") == "72olle"
encrypt_this("good") == "103doo"
encrypt_this("hello world") == "104olle 119drlo"解答：
def encrypt_this(text):
if len(text) == 0:
return text
strList = text.split()
for i,s in enumerate(strList):
if len(s) == 1:
strList[i] = str(ord(s))
elif len(s) == 2:
strList[i] = str(ord(s[0]))+s[1]
else:
strList[i] = str(ord(s[0])) + s[len(s)-1] + s[2:len(s)-1] + s[1]
return ' '.join(strList)4、Sudoku Solution Validator
Sudoku Background
Sudoku is a game played on a 9x9 grid. The goal of the game is to fill all cells of the grid with digits from 1 to 9, so that each column, each row, and each of the nine 3x3 sub-grids (also known as blocks) contain all of the digits from 1 to 9.
(More info at: http://en.wikipedia.org/wiki/Sudoku)
Sudoku Solution Validator
Write a function validSolution/ValidateSolution/valid_solution() that accepts a 2D array representing a Sudoku board, and returns true if it is a valid solution, or false otherwise. The cells of the sudoku board may also contain 0's, which will represent empty cells. Boards containing one or more zeroes are considered to be invalid solutions.
The board is always 9 cells by 9 cells, and every cell only contains integers from 0 to 9.
Examples
validSolution([
[5, 3, 4, 6, 7, 8, 9, 1, 2],
[6, 7, 2, 1, 9, 5, 3, 4, 8],
[1, 9, 8, 3, 4, 2, 5, 6, 7],
[8, 5, 9, 7, 6, 1, 4, 2, 3],
[4, 2, 6, 8, 5, 3, 7, 9, 1],
[7, 1, 3, 9, 2, 4, 8, 5, 6],
[9, 6, 1, 5, 3, 7, 2, 8, 4],
[2, 8, 7, 4, 1, 9, 6, 3, 5],
[3, 4, 5, 2, 8, 6, 1, 7, 9]
]); // => true
validSolution([
[5, 3, 4, 6, 7, 8, 9, 1, 2],
[6, 7, 2, 1, 9, 0, 3, 4, 8],
[1, 0, 0, 3, 4, 2, 5, 6, 0],
[8, 5, 9, 7, 6, 1, 0, 2, 0],
[4, 2, 6, 8, 5, 3, 7, 9, 1],
[7, 1, 3, 9, 2, 4, 8, 5, 6],
[9, 0, 1, 5, 3, 7, 2, 1, 4],
[2, 8, 7, 4, 1, 9, 6, 3, 5],
[3, 0, 0, 4, 8, 1, 1, 7, 9]
]); // => false解答：
# 验证是否包含1-9
def valid(l):
for i in xrange(1,10):
if l.count(i) != 1:
return False
return True
def validSolution(board):
# 验证行
for i in xrange(9):
if not valid(board[i]):
return False
# 验证列
for i in xrange(9):
column = []
for j in xrange(9):
column.append(board[j][i])
if not valid(column):
return False
# 验证3*3阵列
for i in xrange(0,9,3):
for j in xrange(0,9,3):
block = []
block.extend(board[j][i:i+3])
block.extend(board[j+1][i:i+3])
block.extend(board[j+2][i:i+3])
if not valid(block):
return False
return True5、Simple directions reversal
In this Kata, you will be given directions and your task will be to find your way back.
solve(["Begin on Road A","Right on Road B","Right on Road C","Left on Road D"]) = ['Begin on Road D', 'Right on Road C', 'Left on Road B', 'Left on Road A']
solve(['Begin on Lua Pkwy', 'Right on Sixth Alley', 'Right on 1st Cr']) = ['Begin on 1st Cr', 'Left on Sixth Alley', 'Left on Lua Pkwy']解答：
def reverseDirection(str):
if str == 'Left':
return 'Right'
if str == 'Right':
return 'Left'
def solve(arr):
if len(arr) == 1:
return arr
res = []
res.append('Begin' + ' ' + ' '.join(arr[len(arr)-1].split()[1:]))
for i in xrange(len(arr)-1,0,-1):
res.append(reverseDirection(arr[i].split()[0]) + ' ' + ' '.join(arr[i-1].split()[1:]))
return res6、Range Extraction
A format for expressing an ordered list of integers is to use a comma separated list of either
- individual integers
- or a range of integers denoted by the starting integer separated from the end integer in the range by a dash, '-'. The range includes all integers in the interval including both endpoints. It is not considered a range unless it spans at least 3 numbers. For example ("12, 13, 15-17")
Complete the solution so that it takes a list of integers in increasing order and returns a correctly formatted string in the range format.
Example:
solution([-6, -3, -2, -1, 0, 1, 3, 4, 5, 7, 8, 9, 10, 11, 14, 15, 17, 18, 19, 20])
# returns "-6,-3-1,3-5,7-11,14,15,17-20"解答：
def solution(args):
res = [[args[0]]]
for i in range(1, len(args)):
if abs(args[i] - args[i-1]) == 1:
res[len(res)-1].append(args[i])
else:
res.append([args[i]])
for i in range(len(res)):
if len(res[i]) >= 3:
res[i] = str(res[i][0]) + '-' + str(res[i][len(res[i])-1])
elif len(res[i]) == 1:
res[i] = str(res[i][0])
elif len(res[i]) == 2:
res[i] = str(res[i][0]) + ',' +str(res[i][1])
return ','.join(res)7、Positions Average
Suppose you have 4 numbers: '0', '9', '6', '4' and 3 strings composed with them:
s1 = "6900690040"
s2 = "4690606946"
s3 = "9990494604"
Compare s1 and s2 to see how many positions they have in common: 0 at index 3, 6 at index 4, 4 at index 8 ie 3 common positions out of ten.
Compare s1 and s3 to see how many positions they have in common: 9 at index 1, 0 at index 3, 9 at index 5 ie 3 common positions out of ten.
Compare s2 and s3. We find 2 common positions out of ten.
So for the 3 strings we have 8 common positions out of 30 ie 0.2666... or 26.666...%
Given a set of n strings our function pos_average will calculate the average percentage of positions that are the same between the (n * (n-1)) / 2 sets of strings taken amongst the given 'n' strings.
解答：
def pos_average(s):
strList = s.split(', ')
count = 0
if len(strList) == 0 or len(strList) == 1:
return 0
for i in range(len(strList)-1):
for j in range(i+1, len(strList)):
for k in range(len(strList[i])):
if strList[i][k] == strList[j][k]:
count += 1
return count/len(strList)/(len(strList)-1)*2/len(strList[0])*100