题目链接：https://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
bool isMatch(char* s, char* p) {if(*p == '\0' && *s == '\0')return true;if(*p && *(p + 1) == '*') {//如果p后面为*，匹配失败需要回溯if(*p != '.' && *s != *p || *s == '\0') //如果当前字符不匹配,*只能取0，去掉p的首两个字符后递归匹配return isMatch(s, p + 2);if(isMatch(s, p + 2)) //取*为0次，如果能匹配，，返回匹配成功；return true;return isMatch(s + 1, p); //否则回溯，去掉s首字母（*至少匹配了一次）继续递归匹配}else if((*p == '.' && *s != '\0') || *p == *s)//当前字符匹配，递归比较之后序列;return isMatch(s + 1, p + 1);elsereturn false;}//注意'.'不是和所有字符都匹配，因为有可能是'\0';
但我想说，我做了一个善良的平凡女子，并且一直在爱，
