leetcode-wildcard matching-ZZ

http://yucoding.blogspot.com/2013/02/leetcode-question-123-wildcard-matching.html

几个例子：

(1)

acbdeabd

a*c*d

(2)

acbdeabdkadfa

a*c*dfa

Analysis:

For each element in s
If *s==*p or *p == ? which means this is a match, then goes to next element s++ p++.
If p=='*', this is also a match, but one or many chars may be available, so let us save this *'s position and the matched s position.
If not match, then we check if there is a * previously showed up,
       if there is no *,  return false;
       if there is an *,  we set current p to the next element of *, and set current s to the next saved s position.

e.g.

abed
?b*d**

a=?, go on, b=b, go on,
e=*, save * position star=3, save s position ss = 3, p++
e!=d,  check if there was a *, yes, ss++, s=ss; p=star+1
d=d, go on, meet the end.
check the rest element in p, if all are *, true, else false;

Note that in char array, the last is NOT NULL, to check the end, use  "*p"  or "*p=='\0'".

 

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class Solution { public:     bool isMatch(const char *s, const char *p) {         // Start typing your C/C++ solution below         // DO NOT write int main() function                   const char* star=NULL;         const char* ss=s;         while (*s){             if ((*p=='?')||(*p==*s)){s++;p++;continue;}             if (*p=='*'){star=p++; ss=s;continue;}             if (star){ p = star+1; s=++ss;continue;}             return false;         }         while (*p=='*'){p++;}         return !*p;     } };