leetcode - 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         X*X*……都可以跳过 

bool isMatch(const char *s, const char *p) 
{   
    if (s == NULL || p == NULL) return false;
    if (*p == '\0') return *s == '\0';
    // ".*" matches "", so we can't check (*s == '\0') here.
 
    if (*(p + 1) == '*')
    {
        // Here *p != '\0', so this condition equals with
        // (*s != '\0' && (*p == '.' || *s == *p)).
// 
*s != '\0' && *p == '.' 
*p等于'.'时，
若是
*s等于'\0'  则++s超界
// 
*s == *p 则*s不等于'\0' ,所以可以++s；
// 所以只要进入while循环则*s一定不等于'\0',++s一定不会超界
        while ((*s != '\0' && *p == '.') || *s == *p) 
        {
            if (isMatch(s, p + 2)) return true;
            ++s;
        }
        
        return isMatch(s, p + 2);
    }
    else if ((*s != '\0' && *p == '.') || *s == *p)
    {
        return isMatch(s + 1, p + 1);
    }
    
    return false; // *s等于'\0' 或者 *s不等于*P 且没有X*X*之类的可以跳过
}