题解 | #最长公共子序列-II#

python的动态规划实现

#
# longest common subsequence
# @param s1 string字符串 the string
# @param s2 string字符串 the string
# @return string字符串
#
class Solution:
    def LCS(self , s1 , s2 ):
        # write code here
        if len(s1)==0 or len(s2)==0:
            return -1
        m, n = len(s1), len(s2)
        dp = [['']*(n+1) for i in range(m+1)]
        for i in range(1, m+1):
            for j in range(1, n+1):
                if s1[i-1]==s2[j-1]:
                    dp[i][j] = dp[i-1][j-1]+s1[i-1]
                else:
                    dp[i][j] = dp[i-1][j] if len(dp[i-1][j])>=len(dp[i][j-1]) else dp[i][j-1]
        return dp[m][n] if len(dp[m][n])>0 else -1