LeetCode Regular Expression Matching

LeetCode解题之Regular Expression Matching

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原题

简易版正则表达式匹配，只有两种通配符，”.”表示任意一个字符，”c*”表示字符c可以有零个或多个。

注意点：

• 存在一些不合理的字符组合，如”**”
• “.*”可以表示任意字符串
• 需要匹配整个目标串，而不是部分

例子：

输入: s=”aab”, p=”c*a*b”
输出: True

解题思路

开始用递归实现了一遍，结果超时，改用动态规划。dp[i][j]表示s[i:]和p[j:]的匹配情况，围绕”*”进行分类讨论，分类的情况比较多，具体请参看代码注释。

AC源码

class Solution(object):    def isMatch(self, s, p):        """        :type s: str        :type p: str        :rtype: bool        """        m = len(s)        n = len(p)        # Init dp        dp = [[False for i in range(n + 1)] for i in range(m + 1)]        # When string and pattern are all None        dp[m][n] = True        # When the string is None, pattern like "a*" can still match it        for i in range(n - 1, -1, -1):            if p[i] == "*":                dp[m][i] = dp[m][i + 1]            elif i + 1 < n and p[i + 1] == "*":                dp[m][i] = dp[m][i + 1]            else:                dp[m][i] = False        for i in range(m - 1, -1, -1):            for j in range(n - 1, -1, -1):                # When the current character is "*"                if p[j] == "*":                    if j - 1 >= 0 and p[j - 1] != "*":                        dp[i][j] = dp[i][j + 1]                    # If the pattern is starting with "*" or has "**" in it                    else:                        return False                # When the the second character of pattern is "*"                elif j + 1 < n and p[j + 1] == "*":                    # When the current character matches, there are three possible situation                    # 1. ".*" matches nothing                    # 2. "c*" matches more than one character                    # 3. "c*" just matches one character                    if s[i] == p[j] or p[j] == ".":                        dp[i][j] = dp[i][j + 2] or dp[i + 1][j] or dp[i + 1][j + 2]                    # Ignore the first two characters("c*") in pattern since they cannot match                    # the current character in string                    else:                        dp[i][j] = dp[i][j + 2]                else:                    # When the current character is matched                    if s[i] == p[j] or p[j] == ".":                        dp[i][j] = dp[i + 1][j + 1]                    else:                        dp[i][j] = False        return dp[0][0]if __name__ == "__main__":    assert Solution().isMatch("aa", "a") == False    assert Solution().isMatch("aa", "aa") == True    assert Solution().isMatch("aaa", "aa") == False    assert Solution().isMatch("aa", "a*") == True    assert Solution().isMatch("aa", ".*") == True    assert Solution().isMatch("ab", ".*") == True    assert Solution().isMatch("aab", "c*a*b") == True

欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。