最长公共子串和子序列的Python实现，带图示。

使用矩阵来记录两个子串之间各个字符之间的对应关系。

最长子串：矩阵中数字最大的就是最长子串的长度。若对应位置字符相同，则c[i][j] = c[i-1][j-1] + 1

 def longSubStr(str1,str2):
     len1 = len(str1)
     len2 = len(str2)
     longest,start1,start2 = 0,0,0
     c = [[0 for i in range(len2+1)]for i in range(len1+1)]
     for i in range(len1+1):
         for j in range(len2+1):
             if i == 0 or j == 0:
                 c[i][j] = 0
             elif str1[i-1] == str2[j-1]:
                 c[i][j] = c[i-1][j-1]+1
             else:
                 c[i][j] = 0
             if (longest < c[i][j]):
                 longest = c[i][j]
                 start1 = i-longest
                 start2 = j-longest

     return str1[start1:start1+longest],start1,start2

最长子序列：若对应位置字符相同，则c[i][j] = c[i-1][j-1] + 1，若不同，则max(c[i][j-1],c[i-1][j]).

 def printLcs(flag,a,i,j):
     if i==0 or j==0:
         return
     if flag[i][j]=='OK':
         printLcs(flag,a,i-1,j-1)
         print a[i-1],
     elif flag[i][j]=='Left':
         printLcs(flag,a,i,j-1)
     else:
         printLcs(flag,a,i-1,j)  

 def longSubSeq(str1,str2):
     len1 = len(str1)
     len2 = len(str2)
     longest = 0
     c = [[0 for i in range(len2+1)]for i in range(len1+1)]
     flag = [[0 for i in range(len2+1)]for i in range(len1+1)]
     for i in range(len1+1):
         for j in range(len2+1):
             if i == 0 or j == 0:
                 c[i][j] = 0
             elif str1[i-1] == str2[j-1]:
                 c[i][j] = c[i-1][j-1]+1
                 flag[i][j] = 'OK'
                 longest = max(longest,c[i][j])
             elif c[i][j-1] > c[i-1][j]:
                 c[i][j] =c[i][j-1]
                 flag[i][j] = 'Left'
             else:
                 c[i][j] =c[i-1][j]
                 flag[i][j] = 'UP'
     printLcs(flag,str1,len1,len2)
     return longest
 a='ABCBDAB'
 b='BDCABA'
 print longSubSeq(a,b)