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。