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==' '".
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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; }}; |