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

 1 def longSubStr(str1,str2):
 2     len1 = len(str1)
 3     len2 = len(str2)
 4     longest,start1,start2 = 0,0,0
 5     c = [[0 for i in range(len2+1)]for i in range(len1+1)]
 6     for i in range(len1+1):
 7         for j in range(len2+1):
 8             if i == 0 or j == 0:
 9                 c[i][j] = 0
10             elif str1[i-1] == str2[j-1]:
11                 c[i][j] = c[i-1][j-1]+1
12             else:
13                 c[i][j] = 0
14             if (longest < c[i][j]):
15                 longest = c[i][j]
16                 start1 = i-longest
17                 start2 = j-longest
18
19     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]).

 1 def printLcs(flag,a,i,j):
 2     if i==0 or j==0:
 3         return
 4     if flag[i][j]==‘OK‘:
 5         printLcs(flag,a,i-1,j-1)
 6         print a[i-1],
 7     elif flag[i][j]==‘Left‘:
 8         printLcs(flag,a,i,j-1)
 9     else:
10         printLcs(flag,a,i-1,j)
11
12 def longSubSeq(str1,str2):
13     len1 = len(str1)
14     len2 = len(str2)
15     longest = 0
16     c = [[0 for i in range(len2+1)]for i in range(len1+1)]
17     flag = [[0 for i in range(len2+1)]for i in range(len1+1)]
18     for i in range(len1+1):
19         for j in range(len2+1):
20             if i == 0 or j == 0:
21                 c[i][j] = 0
22             elif str1[i-1] == str2[j-1]:
23                 c[i][j] = c[i-1][j-1]+1
24                 flag[i][j] = ‘OK‘
25                 longest = max(longest,c[i][j])
26             elif c[i][j-1] > c[i-1][j]:
27                 c[i][j] =c[i][j-1]
28                 flag[i][j] = ‘Left‘
29             else:
30                 c[i][j] =c[i-1][j]
31                 flag[i][j] = ‘UP‘
32     printLcs(flag,str1,len1,len2)
33     return longest
34 a=‘ABCBDAB‘
35 b=‘BDCABA‘
36 print longSubSeq(a,b)