Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

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

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

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

解题思路

手写最简单的正则表达式函数，判断s是否符合p的正则规则。

初始思路

用两个指针分别读取s和p的头，按照规则依次匹配。确保两个指针可以在满足规则的前提下都移动到s和p的尾端。

进阶思路

动态规划。类似于两个序列做比对，只要可以完全alignment上就OK。

class Solution:
    def isMatch(self, s, p):
        m = len(s) + 1
        n = len(p) + 1
        dp = [[False for _ in range(n)] for _ in range(m)]
        dp[0][0] = True
        for j in range(2, n):
            if p[j-1] == '*':
                dp[0][j] = dp[0][j - 2]
        for r in range(1, m):
            i = r - 1 
            for c in range(1, n):
                j = c - 1 
                if s[i] == p[j] or p[j] == '.':
                    dp[r][c] = dp[r - 1][c - 1]
                elif p[j] == '*':
                    if p[j - 1] == s[i] or p[j - 1] == '.':
                        dp[r][c] = dp[r - 1][c] or dp[r][c - 2]
                    else:
                        dp[r][c] = dp[r][c - 2]
                else:
                    dp[r][c] = False
        return dp[m - 1][n - 1]