1143. 最长公共子序列

class LongestCommonSubsequence2:"""1143. 最长公共子序列https://leetcode.cn/problems/longest-common-subsequence/"""def solution(self, text1: str, text2: str) -> int:"""递归解法 + 备忘录自顶向下:param text1::param text2::return:"""m, n = len(text1), len(text2)self.memo = [[-1 for _ in range(n)] for _ in range(m)]return self.dp(text1, 0, text2, 0)def dp(self, text1, i, text2, j):"""计算text1[i..] 和 text2[j..]的子序列长度:param text1::param i::param text2::param j::return:"""# base caseif i == len(text1) or j == len(text2):return 0# 备忘录if self.memo[i][j] != -1:return self.memo[i][j]if text1[i] == text2[j]:self.memo[i][j] = 1 + self.dp(text1, i+1, text2, j+1)else:self.memo[i][j] = max(self.dp(text1, i + 1, text2, j),self.dp(text1, i, text2, j + 1))return self.memo[i][j]@classmethoddef solution2(cls, text1: str, text2: str) -> int:"""自底向上迭代 动规:param text1::param text2::return:"""m, n = len(text1), len(text2)# dp[m][n] 表示 text1[0..m-1][0..n-1]的lcsdp = [[0 for _ in range(n+1)] for _ in range(m+1)]# base case# dp[0][...]   dp[...][0] = 0for i in range(1, m+1):for j in range(1, n+1):if text1[i-1] == text2[j-1]:dp[i][j] = 1 + dp[i-1][j-1]else:dp[i][j] = max(dp[i][j - 1], dp[i - 1][j])return dp[m][n]