#regular-expression-matching

Given an input string (s) and a pattern (p), 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).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.
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 precedeng 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 (.)".
Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

My code:
class Solution:
    def isMatch(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: bool
        """
        i=0
        j=0
        self.dict={}
        judge=self.check_match(s,p,i,j)
        return bool(judge)
    def check_match(self,s,p,i,j):
        
        if (i,j) in self.dict.keys():
            return self.dict[i,j]
        if j==len(p):
            if i==len(s):
                self.dict[i,j]=1
            else:
                self.dict[i,j]=0
        else:
            match=False
            if i<len(s) and (p[j]==s[i] or p[j]=='.'):
                match=True
            if j+1<len(p) and p[j+1]=='*':
                
                self.dict[i,j]=self.check_match(s,p,i,j+2) or match and self.check_match(s,p,i+1,j)
            else:
                self.dict[i,j]=match and self.check_match(s,p,i+1,j+1)
        return self.dict[i,j]
            
Solution:
https://leetcode.com/problems/regular-expression-matching/