leetcode10_Regular Expression Matching

10. Regular Expression Matching

题目描述：

‘.’ 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

我的代码没成功，思路错了，这个是全匹配。我以为需要p中有s，题目要求p要全匹配s。

简单的python：

import reclass Solution:      # @param {string} s      # @param {string} p      # @return {boolean}      def isMatch(self, s, p):          #一开始理解了半天，后来发现要全匹配，例子最后一个里面的c*相当于没有        return re.match('^' + p + '$', s) != None

复杂点的 leetcode第10题——***Regular Expression Matching：

class Solution(object):      def isMatch(self, s, p):          """         :type s: str         :type p: str         :rtype: bool         """          sLen = len(s)          pLen = len(p)          if (pLen == 0):              return sLen == 0          if (pLen == 1):              if (p == s) or ((p == '.') and (len(s) == 1)):                  return True              else:                  return False          #p的最后一个字符不是'*'也不是'.'且不出现在s里，p跟s肯定不匹配          if (p[-1] != '*') and (p[-1] != '.') and (p[-1] not in s):              return False          if (p[1] != '*'):              if (len(s) > 0) and ((p[0]==s[0]) or (p[0]=='.')):                  return self.isMatch(s[1:],p[1:])              return False          else:              while (len(s) > 0) and ((p[0]==s[0]) or (p[0]=='.')):                  if (self.isMatch(s,p[2:])):                      return True                  s = s[1:]              return self.isMatch(s,p[2:])