题解 | #最长公共子序列-II#

#
# 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
        m, n = len(s1), len(s2)
        
        dp = [[0]*(n+1) for _ 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] + 1
                else:
                    dp[i][j] = max(dp[i][j-1], dp[i-1][j])

        ans = ''
        i, j = m, n 
        while i >= 1 and j >= 1:
            if s1[i-1] == s2[j-1]:
                ans += s1[i-1]
                i -= 1
                j -= 1
            elif dp[i][j-1] > dp[i-1][j]:
                j -= 1
            else: i -= 1
        if not ans: return -1
        return ans[::-1]