10. Regular Expression Matching Leetcode Python

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本文介绍了如何使用动态规划解决正则表达式匹配问题，包括点匹配任意字符和星号匹配零次或多次前导元素。通过实现isMatch函数，详细解释了不同情况下的匹配逻辑，例如空字符串匹配、特定字符重复匹配等。

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

Use DP to solve this problem

code is as follow

class Solution:
    # @return a boolean
    ## have an n*m boolean array
    ## compare with '.' '*' and general case
    ## return dp[-1][-1]
    def isMatch(self, s, p):
        dp = [[False for j in range(len(p) + 1)] for i in range(len(s) + 1)]
        dp[0][0] = True
        for j in range(1,len(p) + 1):
                if p[j-1] == '*':
                    if j >= 2:
                        dp[0][j] = dp[0][j - 2]
        for i in range(1, len(s) + 1):
            for j in range(1, len(p) + 1):
                if p[j - 1] == '.':   ## first case 
                    dp[i][j] = dp[i - 1][j - 1]
                elif p[j - 1] == '*':                                           #aa *a case                     #.* case
                    dp[i][j] = dp[i][j - 1] or dp[i][j - 2] or (dp[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))
                else:
                    dp[i][j] = s[i-1] == p[j-1] and dp[i - 1][j - 1]
        return dp[-1][-1]