## Description

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 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


## Solutions

题意还是比较好理解的，看字串 $p$ 是否能以正则式的形式表达字串 $s$。. 可以表示任意字符，* 表示前一个字符出现 0 到 n 个。

### 1. Recurrence

速度比较快的递归方法：

class Solution(object):
cache = {}
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
if (s, p) in self.cache:
return self.cache[(s, p)]
if not p:
return not s
if p[-1] == '*':
if self.isMatch(s, p[:-2]):
self.cache[(s, p)] = True
return True
if s and (s[-1] == p[-2] or p[-2] == '.') and self.isMatch(s[:-1], p):
self.cache[(s, p)] = True
return True
if s and (p[-1] == s[-1] or p[-1] == '.') and self.isMatch(s[:-1], p[:-1]):
self.cache[(s, p)] = True
return True
self.cache[(s, p)] = False
return False
# Runtime: 28 ms, faster than 93.88% of Python online submissions for Regular Expression Matching.
# Memory Usage: 13.8 MB, less than 5.59% of Python online submissions for Regular Expression Matching.


### 2. DP

找到具体的情况分析，情况比较多！

# Time: O(n^2)
# Space: O(n^2)
class Solution:
def isMatch(self, s: str, p: str) -> bool:
dp = [[False for i in range(len(p) + 1)] for j in range(len(s) + 1)]
dp = True
for i in range(1, len(p) + 1):
if p[i - 1] == '*':
if i >= 2:
dp[i] = dp[i-2]
for i in range(1, len(s) + 1):
for j in range(1, len(p) + 1):
if p[j-1] == '.' or p[j-1] == s[i-1]:
dp[i][j] = dp[i-1][j-1]
elif p[j-1] == '*':
if p[j-2] == s[i-1] or p[j-2] == '.':
dp[i][j] = dp[i][j-2] or dp[i-1][j]
else:
dp[i][j] = dp[i][j-2]
return dp[len(s)][len(p)]

# 447/447 cases passed (52 ms)
# Your runtime beats 60.12 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


其他几种 DP 解法如下：

class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
dp = [[False] * (len(s) + 1) for _ in range(len(p) + 1)]
dp = True
for i in range(1, len(p)):
dp[i + 1] = dp[i - 1] and p[i] == '*'
for i in range(len(p)):
for j in range(len(s)):
if p[i] == '*':
dp[i + 1][j + 1] = dp[i - 1][j + 1] or dp[i][j + 1]
if p[i - 1] == s[j] or p[i - 1] == '.':
dp[i + 1][j + 1] |= dp[i + 1][j]
else:
dp[i + 1][j + 1] = dp[i][j] and (p[i] == s[j] or p[i] == '.')
return dp[-1][-1]
# Runtime: 44 ms, faster than 68.54% of Python online submissions for Regular Expression Matching.
# Memory Usage: 11.7 MB, less than 86.34% of Python online submissions for Regular Expression Matching.

# Time: O(n^2)
# Space: O(n^2)
class Solution:
def isMatch(self, s: str, p: str) -> bool:
dp=[[False for i in range(len(p)+1)] for j in range(len(s)+1)]
dp=True
for i in range(1,len(p)+1):
if p[i-1]=='*':
if i>=2:
dp[i]=dp[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)]

# 447/447 cases passed (48 ms)
# Your runtime beats 68.83 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)