发布于 2015-03-11 03:59:31 | 262 次阅读 | 评论: 0 | 来源: 网友投递
这里有新鲜出炉的Python3 官方中文指南，程序狗速度看过来！
Python编程语言
Python 是一种面向对象、解释型计算机程序设计语言，由Guido van Rossum于1989年底发明，第一个公开发行版发行于1991年。Python语法简洁而清晰，具有丰富和强大的类库。它常被昵称为胶水语言，它能够把用其他语言制作的各种模块（尤其是C/C++）很轻松地联结在一起。
本文实例讲述了Python最长公共子串算法。分享给大家供大家参考。具体如下：
示例代码
#!/usr/bin/env python
# find an LCS (Longest Common Subsequence).
# *public domain*
def find_lcs_len(s1, s2):
m = [ [ 0 for x in s2 ] for y in s1 ]
for p1 in range(len(s1)):
for p2 in range(len(s2)):
if s1[p1] == s2[p2]:
if p1 == 0 or p2 == 0:
m[p1][p2] = 1
else:
m[p1][p2] = m[p1-1][p2-1]+1
elif m[p1-1][p2] < m[p1][p2-1]:
m[p1][p2] = m[p1][p2-1]
else: # m[p1][p2-1] < m[p1-1][p2]
m[p1][p2] = m[p1-1][p2]
return m[-1][-1]
def find_lcs(s1, s2):
# length table: every element is set to zero.
m = [ [ 0 for x in s2 ] for y in s1 ]
# direction table: 1st bit for p1, 2nd bit for p2.
d = [ [ None for x in s2 ] for y in s1 ]
# we don't have to care about the boundery check.
# a negative index always gives an intact zero.
for p1 in range(len(s1)):
for p2 in range(len(s2)):
if s1[p1] == s2[p2]:
if p1 == 0 or p2 == 0:
m[p1][p2] = 1
else:
m[p1][p2] = m[p1-1][p2-1]+1
d[p1][p2] = 3 # 11: decr. p1 and p2
elif m[p1-1][p2] < m[p1][p2-1]:
m[p1][p2] = m[p1][p2-1]
d[p1][p2] = 2 # 10: decr. p2 only
else: # m[p1][p2-1] < m[p1-1][p2]
m[p1][p2] = m[p1-1][p2]
d[p1][p2] = 1 # 01: decr. p1 only
(p1, p2) = (len(s1)-1, len(s2)-1)
# now we traverse the table in reverse order.
s = []
while 1:
print p1,p2
c = d[p1][p2]
if c == 3: s.append(s1[p1])
if not ((p1 or p2) and m[p1][p2]): break
if c & 2: p2 -= 1
if c & 1: p1 -= 1
s.reverse()
return ''.join(s)
if __name__ == '__main__':
print find_lcs('abcoisjf','axbaoeijf')
print find_lcs_len('abcoisjf','axbaoeijf')