Regular Expression Matching

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.

Solution 1:

class Solution:
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        dp = [[False for i in range(len(p)+1)] for j in range(len(s)+1)]
        dp[0][0] = True
        for i in range(1,len(p)+1):
            if p[i-1] == '*':
                if i >= 2:
                    dp[0][i] = dp[0][i-2]
        for i in range(1,len(s)+1):
            for j in range(1,len(p)+1):
                if p[j-1]=='.':
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1]=='*':
                    dp[i][j] = dp[i][j-2] or dp[i][j-1] or (dp[i-1][j] and (s[i-1]==p[j-2] or p[j-2]=='.'))
                else:
                    dp[i][j] = dp[i-1][j-1] and s[i-1] == p[j-1]
        return dp[len(s)][len(p)]   

Solution 2:

class Solution:
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        if re.compile('^'+p+'$').match(s):
            return True
        return False