Python 最长公共子串算法，python串算法,[Python]代码#!

Python 最长公共子串算法，python串算法,[Python]代码#!

[Python]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')