Leetcode 10. 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 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:
# X X X X X
# X X X *
m = len(s)
n = len(p)
dp = [ for _ in range(m + 1)]
s = '#' + s
p = '#' + p
dp = True
for j in range(2, n + 1):
dp = (p == '*' and dp)
for i in range(1, m + 1):
for j in range(1, n + 1):
if p.isalpha():
dp = (s == p and dp)
elif p == '.':
dp = dp
elif p == '*':
dp = ((s == p or p == '.') and dp) or dp
return dp
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