day9

看到答案和我的代码以后，相形惭愧啊

 1 def is_magic_word(s):
 2     i = 0
 3     index = 0
 4     while i < len(s) - 1:
 5         if s[i] == s[i+1]:
 6             index = i
 7             break
 8         i += 1
 9     if (index+5) < len(s):
10         if s[index+2] != s[index+3] or s[index+4] != s[index+5]:
11             return False
12         else:
13             return True
14     else:
15         return False
16         
17 file = open("words.txt")
18 
19 for line in file:
20     word = line.strip()
21     if is_magic_word(word):
22         print(word)
 1 """This module contains a code example related to
 2 
 3 Think Python, 2nd Edition
 4 by Allen Downey
 5 http://thinkpython2.com
 6 
 7 Copyright 2015 Allen Downey
 8 
 9 License: http://creativecommons.org/licenses/by/4.0/
10 """
11 
12 from __future__ import print_function, division
13 
14 
15 def is_triple_double(word):
16     """Tests if a word contains three consecutive double letters.
17     
18     word: string
19 
20     returns: bool
21     """
22     i = 0
23     count = 0
24     while i < len(word)-1:
25         if word[i] == word[i+1]:
26             count = count + 1
27             if count == 3:
28                 return True
29             i = i + 2
30         else:
31             count = 0
32             i = i + 1
33     return False
34 
35 
36 def find_triple_double():
37     """Reads a word list and prints words with triple double letters."""
38     fin = open('words.txt')
39     for line in fin:
40         word = line.strip()
41         if is_triple_double(word):
42             print(word)
43 
44 
45 print('Here are all the words in the list that have')
46 print('three consecutive double letters.')
47 find_triple_double()
48 print('')

 

 

再扔一段我的丑陋代码

 1 def is_huiwen(num):
 2     s = str(num)
 3     if len(s) != 6:
 4         print('wrong')
 5     else:
 6         if s[2] == s[5] and s[3] == s[4]:
 7             return True
 8         if s[1] == s[5] and s[2] == s[4]:
 9             return True
10         if s[0] == s[5] and s[1] == s[4] and s[2] == s[3]:
11             return True
12         return False
13 count = 0        
14 for i in range(100000,1000000):
15     if is_huiwen(i):
16         count += 1
17         print(i)
18 print(count)

 

感觉被习题的最后一题完全带跑偏了...

 1 def get_mom_age(age_son):
 2     s = str(age_son)
 3     s1 = s[::-1]
 4     age_mom = int(s1)
 5     return age_mom
 6     
 7 # print(get_mom_age(67))
 8     
 9 def is_fun(age_son, age_mom):
10     if age_mom == get_mom_age(age_son):
11         return True
12     else:
13         return False
14 
15 def is_fun_num(age_son):
16     count = 0
17     age_mom = get_mom_age(age_son)
18     
19     new_age_son = age_son
20     new_age_mom = age_mom
21     
22     while new_age_son >= 1:
23         new_age_son = new_age_son - 1
24         new_age_mom = new_age_mom - 1
25         if is_fun(new_age_son, new_age_mom):
26             # print(new_age_son, new_age_mom)
27             count += 1
28         else:
29             pass
30         
31     if count == 6:
32         new_age_mom = age_mom
33         new_age_son = age_son
34         
35         while new_age_son < 100:
36             new_age_son = new_age_son + 1
37             new_age_mom = new_age_mom + 1
38             if is_fun(new_age_son, new_age_mom):
39                 # print(new_age_son, new_age_mom)
40                 count += 1
41             else:
42                 pass
43                 
44         if count == 8:
45             return True
46         else:
47             return False
48     else:
49         return False
50             
51 for i in range(10,100):
52     s = str(i)
53     if s[0] > s[1]:
54         # print('no way', i)
55         pass
56     else:
57         if is_fun_num(i):
58             print(i)

 今天看到按值传递和按引用传递了，传一个列表给函数是传一个引用的，这样意味在函数内对列表的修改都会在函数外部有效。

我们正常传一个数值类的值或字符串的时候，都不用考虑这个问题，都可以当成是函数怎么对参数修改，都不会影响函数外这个参数的值。就相当与按值传递，也就是说复制了一份值。

有些太晚了，不要去太深入思考这个问题了，容易睡不着觉，今天还是浪费了比较多的时间，后续要提高提高效率。

posted @ 2017-06-21 23:27  onhacker  阅读(167)  评论(0)    收藏  举报