Python最长公共子串算法实例

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：

#!/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')

希望本文所述对大家的Python程序设计有所帮助。