【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") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("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+; 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;
}
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