# Regular Expression Matching

Given an input string `s` and a pattern `p`, implement regular expression matching with support for `'.'` and `'*'` where:

• `'.'` Matches any single character.​​​​
• `'*'` Matches zero or more of the preceding element.

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

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 (.)".
```

Constraints:

• `1 <= s.length <= 20`
• `1 <= p.length <= 30`
• `s` contains only lowercase English letters.
• `p` contains only lowercase English letters, `'.'`, and `'*'`.
• It is guaranteed for each appearance of the character `'*'`, there will be a previous valid character to match.

### Solution 1:

``````class Solution:
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
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-2] or dp[i][j-1] 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)]   ``````

### Solution 2:

``````class Solution:
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
if re.compile('^'+p+'\$').match(s):
return True
return False ``````