【科学刷题】模式匹配专题

44. 通配符匹配

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        # 忘了
        dp[0][0] = True
        for j in range(1, n + 1):
            # * 能和 空串 匹配
            if p[j - 1] == '*':
                dp[0][j] = True
            else:
                break
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j - 1] == '*':
                    #         空串情况         上一个情况
                    dp[i][j] = dp[i][j - 1] | dp[i - 1][j]
                else:
                    if p[j - 1] == '?' or s[i - 1] == p[j - 1]:
                        dp[i][j] |= dp[i - 1][j - 1]
        return dp[m][n]