class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
# 正确初始化 DP table
dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
# 注意循环范围,应该从 1 开始,因为 dp[i][j] 对应 text1[:i] 和 text2[:j]
for i in range(1, m + 1):
for j in range(1, n + 1):
# 注意字符串索引要减 1,因为字符串索引从 0 开始
if text1[i - 1] == text2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
# 结果在 dp 表的右下角
return dp[m][n] # 或者 dp[-1][-1] 也可以