Leetcode: Wildcard Matching

最新推荐文章于 2019-04-04 14:47:25 发布

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最新推荐文章于 2019-04-04 14:47:25 发布

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分类专栏：算法与数据结构程序员笔试面试 Leetcode 文章标签： Leetcode Wildcard Matching 递归

本文链接：https://blog.youkuaiyun.com/doc_sgl/article/details/12721187

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本文介绍如何实现支持'*'和'?'通配符的模式匹配算法。通过遍历字符串`s`，判断每个字符与模式`p`中的字符是否匹配。如果遇到'*'，则可能匹配一或多个字符，记录当前位置并继续。若不匹配，检查之前是否存在'*'，没有则返回false，有则更新`s`和`p`的位置。最后，检查`p`中剩余元素是否全为'*'，若是则返回true，否则返回false。

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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

递归： Time Limit Exceeded

bool isMatch(const char *s, const char *p) {
        // Note: The Solution object is instantiated only once.    
        if(*s == '\0')
		{
			if((*p == '\0') || (*p == '*' && *(p+1) == '\0'))
				return true;
			else
				return false;
		}
		if(*p == '*')
		{
			while(*(p+1) == '*') p++;
			return isMatch(s, p+1) || isMatch(s+1, p);
		}
		else if((*p == '?') || (*s == *p))
		{
			while(*s != '\0' && ((*p == '?') || (*s == *p)))
			{
				s++;
				p++;
			}
			return isMatch(s, p);
		}
		else
			return false;
    }

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'".

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;
    }

ref:

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