地址：https://leetcode.com/problems/regular-expression-matching/description/
描述：
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
递归
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
#递归
if len(p)==0:
return len(s)==0
if (len(p)==1) or (p[1]!='*'):
if len(s)==0 or (s[0]!=p[0] and p[0]!='.'):
return False
return self.isMatch(s[1:], p[1:])
else:
i = -1
length = len(s)
while i<length and (i<0 or p[0]=='.' or p[0]==s[i]):
if self.isMatch(s[i+1:], p[2:]):
return True
i += 1
return False
动态规划
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
#动态规划
dp = [[False for i in range(len(p)+1)] for j in range(len(s)+1)]
dp[0][0] = True
for i in range(1, len(p)+1):
if p[i-1]=='*':
if i>=2:
dp[0][i]=dp[0][i-2]
for i in range(1,len(s)+1):
for j in range(1,len(p)+1):
if p[j-1]=='.':
dp[i][j]=dp[i-1][j-1]
elif p[j-1]=='*':
dp[i][j]=dp[i][j-1] or dp[i][j-2] or (dp[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))
else:
dp[i][j]=dp[i-1][j-1] and s[i-1]==p[j-1]
return dp[len(s)][len(p)]
