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