10 Regular Expression Matching

https://leetcode.com/problems/regular-expression-matching/

solution

• 动态规划

  • * 意味着从前面转移而来

# dp[i][j]: s[0:i-1], p[0: j-1]的子字符串是否符合

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m = len(s)
        n = len(p)
        dp = [[False] * (n + 1) for _ in range(m + 1)]

        dp[0][0] = True
        for j in range(2, n + 1):
            if p[j-1] == '*':
                dp[0][j] = dp[0][j-2]  # *取0个时

        for i in range(1, m+1):
            for j in range(1, n+1):
                if s[i-1] == p[j-1] or p[j-1] == '.':
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1] == '*':
                    if p[j-2] == s[i-1] or p[j-2] == '.':
                        dp[i][j] = dp[i][j-2] or dp[i-1][j]
                    else:
                        dp[i][j] = dp[i][j-2]
        return dp[-1][-1]

时间复杂度：O(mn) 空间复杂度：O(mn)

follow up

44 Wildcard Matching