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【leetcode】Wildcard Matching（hard) ★ 大神太牛了
Implement wildcard pattern matching with support for ‘?‘ and ‘*‘.
‘?‘ Matches any single character.‘*‘ Matches any sequence of characters (including the empty sequence).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", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false
我的思路：不提了，太挫了，写了100多行代码都没搞定，直接看大神10行搞定的代码吧：
其实主要的问题就在于p中的*究竟代表哪几个字母，大神的代码中用ss记录*代表字母的后面一个位置，star记录p中*的位置。
先假设*代表0个字符，如果后面发现不成立，再返回来
设*代表1个字符，............................
bool isMatch(const char *s, const char *p) { const char* star=NULL; const char* ss=s; while (*s){ //advancing both pointers when (both characters match) or (‘?‘ found in pattern) //note that *p will not advance beyond its length if ((*p==‘?‘)||(*p==*s)){s++;p++;continue;} // * found in pattern, track index of *, only advancing pattern pointer if (*p==‘*‘){star=p++; ss=s;continue;} //current characters didn‘t match, last pattern pointer was *, current pattern pointer is not * //only advancing pattern pointer if (star){ p = star+1; s=++ss;continue;} //current pattern pointer is not star, last patter pointer was not * //characters do not match return false; } //check for remaining characters in pattern while (*p==‘*‘){p++;} return !*p; }
【leetcode】Wildcard Matching（hard) ★ 大神太牛了
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