LeetCode Regular Expression Matching

LeetCode Regular Expression Matching
LeetCode-solving Regular Expression Matching

Original question

The simple regular expression matches with only two wildcards. "." indicates any character, and "c *" indicates that the character c can have zero or multiple characters.

Note:

There are some unreasonable character combinations, such as "**" ". *", which can indicate that any string needs to match the entire target string, rather than some

Example:

Input: s = "aab", p = "c * a * B"
Output: True

Solutions

It is implemented recursively and the result times out. Instead, it uses dynamic planning. Dp [I] [j] indicates the matching between s [I:] and p [j:]. The classification is discussed around "*". There are many categories, for more information, see code comments.

AC Source Code

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