44. Wildcard Matching
Given an input string (s
) and a pattern (p
), implement wildcard pattern matching with support for '?'
and '*'
.
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
Note:
s
could be empty and contains only lowercase lettersa-z
.p
could be empty and contains only lowercase lettersa-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 = "*"
Output: true
Explanation: '*' matches any sequence.
Example 3:
Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.
Example 4:
Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
Example 5:
Input:
s = "acdcb"
p = "a*c?b"
Output: false
Code:
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
l = len(s)
if len(p) - p.count('*') > l:
return False
dp = [True] + [False] * l
for letter in p:
new_dp = [dp[0] and letter == '*']
if letter == '*':
for j in range(l):
new_dp.append(new_dp[-1] or dp[j + 1])
elif letter == '?':
new_dp += dp[:l]
else:
new_dp += [dp[j] and s[j] == letter for j in range(l)]
dp = new_dp
return dp[-1]