*10. Regular Expression Matching [Hard] 字符串匹配/DP/递归

最新推荐文章于 2022-04-21 15:48:07 发布

木木歹朱

最新推荐文章于 2022-04-21 15:48:07 发布

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10. Regular Expression Matching

class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        if len(p)==0:
            return len(s)==0
        elif len(p)==1:
            return len(s)==1 and ( p=='.' or p==s )
        elif p[1]!='*':
            if len(s)==0:
                return False
            return (s[0]==p[0] or p[0]=='.') and self.isMatch(s[1:],p[1:])
        else:
            while len(s)>0 and ( s[0]==p[0] or p[0]=='.' ):
                if self.isMatch(s,p[2:]):
                    return True
                s = s[1:]
            return self.isMatch(s,p[2:])

class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        m, n = len(s), len(p)
        dp = [[False for i in range(n+1)] for j in range(m+1)]
        dp[0][0] = True
        for i in range(1, n + 1):
            if p[i-1] == '*':
                dp[0][i] = dp[0][i-2]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j-1] == '*':
                    if s[i-1] == p[j-2] or p[j-2] == '.':
                        dp[i][j] = dp[i][j-2] or dp[i][j-1] or dp[i-1][j] 
                    else:
                        dp[i][j] = dp[i][j-2]
                else:
                    dp[i][j] = (s[i-1] == p[j-1] or p[j-1] == '.') and dp[i-1][j-1]
        return dp[m][n]