# LeetCode 10：Regular Expression Matching

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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true```

The matching should cover the entire input string (not partial).

```class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
import re
pattern = '^%s\$' % p
return re.search(pattern, s) is not None```

https://www.jianshu.com/p/85f3e5a9fcda

Java实现源码也来自上面链接

```package com.maoxiaomeng;

/**
* @author lihui
* @date 2018/3/26 12:31
*/

class Solution {
public boolean isMatch(String s, String p) {
if (p.isEmpty()) {
return s.isEmpty();
}
if (p.length() == 1 || p.charAt(1) != '*') {
if (s.isEmpty() || (p.charAt(0) != '.' && p.charAt(0) != s.charAt(0))) {
return false;
} else {
return isMatch(s.substring(1), p.substring(1));
}
}
while (!s.isEmpty() && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.')) {
if (isMatch(s, p.substring(2))) {
return true;
}
s = s.substring(1);
}
return isMatch(s, p.substring(2));
}
}```