10. Regular Expression Matching
Given an input string (s) and a pattern §, 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
仔细看examples,注意’*‘的两种情况，一个是代表前一个字符重复0次（相当于忽略前一个字符），另一个是代表前一个字符重复n次，而且’*‘是不可以单独出现的；
感觉使用递归比动态规划更容易理解一些；
返回条件是p为空的时候，子问题是对’*‘的情况的分类考虑；
可以首先考虑没有’*'的情况，比较简单，就判断当前s[0]和p[0]是否匹配以及子问题是否匹配，即：
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
if not p:
return not s
first_m = bool(s) and p[0] in {s[0], '.'}
return first_m and self.isMatch(s[1:], p[1:])
加上’*‘就是考虑当p[2]==’*'的时候的两种情况，要么直接忽略前一个字符，要么当前匹配，而且重复n次：
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
if not p:
return not s
first_m = bool(s) and p[0] in {s[0], '.'}
if len(p)>=2 and p[1] == '*':
return (self.isMatch(s, p[2:])) or (first_m and self.isMatch(s[1:], p))
else:
return first_m and self.isMatch(s[1:], p[1:])