#10 Regular Expression Match

标签：leetcode   string   backtracking   c   

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';

#10 Regular Expression Match

标签：leetcode   string   backtracking   c   

原文地址：http://blog.csdn.net/ice_camel/article/details/46582423