题解 | #最长公共子序列(二)#
最长公共子序列(二)
https://www.nowcoder.com/practice/6d29638c85bb4ffd80c020fe244baf11
import enum # # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # 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: str, s2: str) -> str: # write code here n, m = len(s1), len(s2) f = [[0] * (m + 1) for _ in range(n + 1)] for i, x in enumerate(s1): for j, y in enumerate(s2): if x == y: f[i + 1][j + 1] = f[i][j] + 1 else: f[i + 1][j + 1] = max(f[i][j + 1], f[i + 1][j]) if f[-1][-1] == 0: return str(-1) ans, i, j = '', n, m while i > 0 and j > 0: if s1[i - 1] == s2[j - 1]: # 找到了 ans += s1[i - 1] i -= 1 j -= 1 else: if f[i][j - 1] >= f[i - 1][j]: # [i, j-1] 转移过来的 j -= 1 else: i -= 1 return ans[::-1]