10. Regular Expression Matching leetcode Python 2016 new Season-优快云博客

本文链接：https://blog.youkuaiyun.com/hyperbolechi/article/details/50458341

本文详细介绍了如何实现正则表达式的匹配功能，支持使用'*'匹配零次或多次前一元素，及使用'.'匹配任意单字符。提供了函数原型和示例说明。

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Implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

This problem can be solved with O(n^2) time and space algorithm

1. letter matching

2. letter match with * case

3. letter match with . case

class Solution(object):
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        length_s =  len(s)
        length_p = len(p)
        dp = [[False for _ in range(length_p + 1)] for _ in range(length_s + 1)]
        dp[0][0] = True
        for column in range(1, length_p + 1):
            if p[column - 1] == '*' and column >= 2:
                dp[0][column] = dp[0][column - 2]
        for row in range(1, length_s + 1):
            for column in range(1, length_p + 1):
                if p[column - 1] == '.':
                    dp[row][column] = dp[row - 1][column - 1]
                elif p[column - 1] == '*':                                            # aa *a      aa .*
                    dp[row][column] = dp[row][column - 1] or dp[row][column - 2] or (dp[row - 1][column] and (s[row - 1] == p[column - 2] or p[column - 2] == '.'))
                else:
                    dp[row][column] = s[row - 1] == p[column - 1] and dp[row - 1][column - 1]
        return dp[-1][-1]