Python longest common substring algorithm instance

This article mainly introduces the longest public substring algorithm in Python, and analyzes the skills of Python string operations. if you need it, refer to the example in this article to describe the longest public substring algorithm in Python. Share it with you for your reference. The details are as follows:

#!/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')

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