codewars练习

1. 如果我们列出所有小于10的自然数中，3或5的倍数，我们可以得到3 5 6 9。这几个数的和为23。

#给一个数，返回所有小于它的自然数，3和5的倍数之和。如果其中有负数，则返回0。

# 样例：solution(4), 3 ，solution(6), 8。

 1 def solution(number):
 2     sum = 0
 3     if number<0:
 4         return sum
 5     else:
 6         for i in range(1,number):
 7             if i % 3 ==0 or i% 5 == 0:
 8                 sum=sum+i
 9         return sum
10

2. 重构不区分大小写出现一次的字符为"(",出项两次及以上的为")"

# 样例: "din" => "((("    "recede" => "()()()"

 1 from collections import Counter
 2 def duplicate_encode(word):
 3     dic1 = Counter(word.lower())
 4     new_word = list(word.lower())
 5     for i in range(len(new_word)):
 6         if dic1[new_word[i]]>1:
 7             new_word[i] =')'
 8         else:
 9             new_word[i] = '('
10     return ''.join(new_word)
11 
12 def duplicate_encode(word):
13     count={}
14     for i in word.lower():
15         if i not in count:
16             count[i]=1
17         else:
18             count[i]+=1
19     print(count)
20     new_word=word.lower()
21     s=""
22     for i in new_word:
23         if count[i]>1:
24             s+=')'
25         else:
26             s+='('
27     return s

3. Create Phone Number

# 样例[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]), "(123) 456-7890"
def create_phone_number2(n):
    str1 =  ''.join(str(x) for x in n[0:3])
    str2 =  ''.join(str(x) for x in n[3:6])
    str3 =  ''.join(str(x) for x in n[6:10])
    return '({}) {}-{}'.format(str1, str2, str3)
def create_phone_number1(n):
    return "({}{}{}) {}{}{}-{}{}{}{}".format(*n)

def create_phone_number(n):
    #your code here
    s = ""
    m = ""
    for i in n:
        if len(s)<=2:
            s += str(i)
    for i in range(len(n)):
        if i>=3:
            m += str(n[i])
            if i==5:
                 m += "-"
    return("("+s+")"+" "+m)
create_phone_number2([0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

4. Are they the "same"?

# 判断数组一的平方是否为数组二

# 样例 a1 = [-121, -144, 19, -161, 19, -144, 19, -11] a2 = [121, 14641, 20736, 361, 25921, 361, 20736, 361]

def comp(array1, array2):
    # your code
    count = 0
    if(array1 == None) or (array2 == None) or(len(array1)!=len(array2)):
        return False
    length = len(array1)
    for i in range(length):
        if array1[i]<0:
            array1[i] = -array1[i]
    data_1 = sorted(array1) 

    data_2 = sorted(array2) 
    print(data_1)
    print(data_2)
    for i in range(length):
        if pow(data_1[i],2) == data_2[i]:
            count=count+1
    if count==length:
        return True
    else:
        return False

5. Mutual Recursion

# F(0) = 1; M(0) = 0; F(n) = n - M(F(n - 1)); M(n) = n - F(M(n - 1))

def f(n):
    if n==0:
        return 1
    return n - m(f(n - 1))
def m(n):
    if n==0:
        return 0
    return n - f(m(n - 1))

6. Count the smiley faces!

 1 # face marked as : or ; 
 2 # face can have a nose but it does not have to.  a nose are - or ~
 3 # face must have a smiling mouth,marked with either ) or D
 4 # 样例: [':D',':',';~D',':)'], 4
 5 def count_smileys(arr): # arr(eyes, nose, mouth)
 6     num = 0
 7     print(arr)
 8     if arr is None:
 9         return num
10     else:
11         for i in range(len(arr)):
12             if len(arr[i]) == 3:
13                 if arr[i][0] in [":",";"] and arr[i][-1] in [")","D"] and arr[i][1] in ["-","~",""]:
14                     num+=1 
15                 else:
16                     num += 0
17             else:
18                 if arr[i][0] in [":",";"] and arr[i][-1] in [")","D"] :
19                     num+=1 
20                 else:
21                     num += 0
22         return num
23 count_smileys([':oD', ':D', ';oD', ':(', ':(', ';oD', ';-(', ':oD', ':-(', ';(', ':D', ';D'])

7. Consonant value

 1 # We shall assign the following values: a = 1, b = 2, c = 3, .... z = 26.
 2 #  Consonants are any letters of the alphabet except "aeiou".
 3 #  For example, for the word "zodiacs", let's cross out the vowels. We get: "z  d  cs"
 4 # z = 26, d = 4 and cs = 3 + 19 = 22. The highest is 26.
 5 alphabet='abcdefghijklmnopqrstuvwxyz'
 6 alphabet_dict={}
 7 for i in range(len(alphabet)):
 8     alphabet_dict[alphabet[i]]=i+1
 9 def solve(s):
10 #     for i in s:
11 #         if i in ["a","e","i","o","u"]:
12 #             s = s.replace(i,"1")
13     s = s.replace('a','1').replace('e','1').replace('i','1').replace('u','1').replace('o','1')
14     s = s.split("1")
15     a = list(filter(None, s))  # 只能过滤空字符和None
16     list_a=[]
17     for i in a:
18         sum = 0
19         for j in range(len(i)):
20             sum+=alphabet_dict[i[j]]
21             j+=1
22         list_a.append(sum)
23     print(max(list_a)) 
24 solve("catchphrase")

posted on 2021-10-31 14:22 GooBo 阅读(110) 评论(0) 收藏举报