[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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

===Comments by Dabay===自我感觉很解得很垃圾啊。如果不是把p做了压缩还过不了online judge。其实就是一个DFA。

当s和p都是空的时候，匹配成功。如果s为空，p不为空，检查p的偶数为是不是都是*。

如果s和p都不为空，头一个字母    如果不匹配，        不带*，False        带*，递归p[2:]    如果匹配，        不带*，递归s[1:],p[1:]        带*，考虑匹配0到最远端的情况，分别递归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)