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") → 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
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); else return false; } //注意'.'不是和所有字符都匹配,因为有可能是'\0';
时间: 2024-10-26 10:35:09