Question

我是Python的新手，目前正在阅读“潜入Python”中的字符串操作章节。

我想知道有哪些最佳（或最聪明/最有创意）的方法可以做到以下几点：

1）从这个字符串中提取：“stackoverflow.com/questions/ask”这个词'问题'。我做了string.split（/）[0] - 但这不是很聪明。

2）找到给定数字或字符串中最长的回文

3）从一个给定的单词（即“cat”）开始 - 找到所有可能的方法从那里获得另一个三个字母的单词（“dog”），一次更改一个字母，以便每次更改字母形成一个新的有效词。

例如 - 猫，婴儿床，点，狗

Answer 1

作为个人练习，这是给你的，（希望）很好的评论代码和一些提示。

#!/usr/bin/env python2

# Let's take this string:
a = "palindnilddafa"
# I surround with a try/catch block, explanation following
try:
  # In this loop I go from length of a minus 1 to 0.
  # range can take 3 params: start, end, increment
  # This way I start from the thow longest subsring,
  # the one without the first char and without the last
  # and go on this way
  for i in range(len(a)-1, 0, -1):
    # In this loop I want to know how many 
    # Palidnrome of i length I can do, that
    # is len(a) - i, and I take all
    # I start from the end to find the largest first
    for j in range(len(a) - i):
      # this is a little triky.
      # string[start:end] is the slice operator
      # as string are like arrays (but unmutable).
      # So I take from j to j+i, all the offsets 
      # The result of "foo"[1:3] is "oo", to be clear.
      # with string[::-1] you take all elements but in the
      # reverse order
      # The check string1 in string2 checks if string1 is a 
      # substring of string2
      if a[j:j+i][::-1] in a:
        # If it is I cannot break, 'couse I'll go on on the first
        # cycle, so I rise an exception passing as argument the substring
        # found
        raise Exception(a[j:j+i][::-1])

# And then I catch the exception, carrying the message
# Which is the palindrome, and I print some info
except Exception as e:
  # You can pass many things comma-separated to print (this is python2!)
  print e, "is the longest palindrome of", a
  # Or you can use printf formatting style
  print "It's %d long and start from %d" % (len(str(e)), a.index(str(e)))

讨论结束后，如果它发生了，我很抱歉。我已经编写了另一个回文搜索器的实现，如果sberry2A可以，我想知道一些基准测试的结果！

请注意，关于指针和硬“+1 -1”问题存在很多错误（我猜），但这个想法很清楚。从中间开始，然后展开，直到可以。

以下是代码：

#!/usr/bin/env python2


def check(s, i):
  mid = s[i]
  j = 1
  try:
    while s[i-j] == s[i+j]:
      j += 1
  except:
    pass
  return s[i-j+1:i+j]

def do_all(a):
  pals = []
  mlen = 0
  for i in range(len(a)/2):
    #print "check for", i
    left = check(a, len(a)/2 + i)
    mlen = max(mlen, len(left))
    pals.append(left)

    right = check(a, len(a)/2 - i)
    mlen = max(mlen, len(right))
    pals.append(right)

    if mlen > max(2, i*2-1):
      return left if len(left) > len(right) else right

string = "palindnilddafa"

print do_all(string)

Answer 2

No 3：

如果您的字符串为s：

max((j-i,s[i:j]) for i in range(len(s)-1) for j in range(i+2,len(s)+1) if s[i:j]==s[j-1:i-1:-1])[1]

将返回答案。

Answer 3

＃2 - How to find the longest palindrome in a given string?

Python：字符串操作的聪明方法

3 个答案: