公共子串问题

在计算机科学中，最长公共子串问题是寻找两个或多个已知字符串最长的子串。此问题与最长公共子序列问题的区别在于子序列不必是连续的，而子串却必须是。链接: 百度百科

费曼算法

The Feynman Algorithm:

1. Write down the problem.
2. Think real hard.
3. Write down the solution.

动态规划算法思路

1. 给定两个字符串T1和T2, 假设dp(i, j)表示T1，T2在位置i, j的公共子串的长度【向位置0的方向，连续相同的字符的数量】
   比如“abcdefg”, “absecde”
   dp(0,0) = 1 ‘a’ == ‘a’
   dp(1,1) = 2 ‘ab’ == ‘ab’
   dp(1,2) = 0 ‘b’ != ‘s’

2. 总结规律：
   if T1[i] == T2[j] :
   dp(i,j) = dp(i-1,j-1) + 1
   else:
   dp(i,j) = 0

代码实现

def getLcs(t1:str, t2:str)->str:'''get the longest common substring of two given string'''if not t1:return t1if not t2:return t2# dp = [[0 for i in range(len(t2)+1)] for j in range(len(t1)+1)]max_len = 0dp = [[0 for i in range(len(t2)+1)] for j in range(2)]for i in range(1, len(t1)+1):ind = [0,1][i%2==0]for j in range(1, len(t2)+1):if t1[i-1]==t2[j-1]:dp[ind][j] = dp[1-ind][j-1] + 1else:dp[ind][j] = 0if dp[ind][j] >max_len:max_len = dp[ind][j]p = iprint("max_len:%d , pos: %d"%(max_len,p))return t1[p-max_len:p]