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 -
scontains only lowercase English letters. -
pcontains only lowercase English letters,'.', and'*'. - It is guaranteed for each appearance of the character
'*', there will be a previous valid character to match.
SOLUTION:
class Solution:
def isSame(self, a, b):
if a == "." or b == ".":
return True
return a == b
def isMatchRec(self, s, p, i, j, m, n):
if i >= m:
if j >= n:
return True
return j < n - 1 and p[j + 1] == "*" and self.isMatchRec(s, p, i, j + 2, m, n)
if j >= n:
return i >= m
if j == n - 1 or p[j + 1] != "*":
return self.isSame(s[i], p[j]) and self.isMatchRec(s, p, i + 1, j + 1, m, n)
if p[j + 1] == "*":
return self.isMatchRec(s, p, i, j + 2, m, n) or (self.isSame(s[i], p[j]) and self.isMatchRec(s, p, i + 1, j, m, n))
return False
def isMatch(self, s: str, p: str) -> bool:
m = len(s)
n = len(p)
return self.isMatchRec(s, p, 0, 0, m, n)
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