[Leetcode] [Python] Regular Expression Matching

#-*-Coding:utf8-*-
‘‘‘
https://oj.leetcode.com/problems/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

===comments by dabay===
I feel so much rubbish about myself. If I did not compress p, I could not get through online judge. is actually a DFA.

When S and P are both empty, the match succeeds.
If S is empty, p is not empty, check the even number of P is not all *.

If S and p are not empty, the first letter
If they do not match,
With no *,false
with *, recursive p[2:]
If it matches,
Without *, recursive s[1:],p[1:]
with *, consider matching 0 to the far end of the case, respectively recursive s[i:],p[2:]
‘‘‘
Class Solution:
# @return A Boolean
def isMatch (self, S, p):
def compress (p):
i = 0
While I < Len (p)-3:
If p[i+1]! = ' * ':
i = i + 1
Continue
If p[i+3] = = ' * ':
If p[i] = = "." or p[i+2] = = ".":
p = p[:i] + ". *" + p[i+4:]
Continue
Elif P[i] = = P[i+2]:
p = p[:i] + p[i+2:]
Continue
i = i + 2
Return P

def ISMATCH2 (S, p):
If Len (s) = = 0:
If Len (p) = = 0:
Return True
If Len (p)% 2 = = 0:
i = 1
While I < Len (p):
If p[i]! = "*":
Return False
i = i + 2
Else
Return True
Else
Return False
If Len (s) > 0 and len (p) = = 0:
Return False

Match_char = p[0]
Multi = False
If Len (P) > 1:
If p[1] = = "*":
Multi = True

If Match_char! = S[0] and Match_char! = ".":
If multi:
Return IsMatch2 (S, p[2:])
Else
Return False

If Multi is False:
Return IsMatch2 (s[1:], p[1:])
Else
result = False
i = 0
result = ISMATCH2 (S, p[2:])
While I < Len (s):
if result = = True:
return result
if (s[i] = = Match_char or Match_char = = "."):
result = ISMATCH2 (s[i+1:], p[2:])
Else
Break
i = i + 1
return result

Return ISMATCH2 (S, Compress (p))


def main ():
s = solution ()
Print S.ismatch ("AA", "A *")


if __name__ = = "__main__":
Import time
Start = Time.clock ()
Main ()
Print "%s sec"% (Time.clock ()-start)

[Leetcode] [Python] Regular Expression Matching