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 lettersa-z
.p
could be empty and contains only lowercase lettersa-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[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] == '.' 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[0][0] = True
for i in range(1, len(p)):
dp[i + 1][0] = dp[i - 1][0] 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[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)]
# 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)