Wildcard Matching 字符串含？，*匹配

Wildcard Matching

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

class Solution {
public:

/*
//递归解法，会TLE。
    bool isMatch(string s, string p) {
        
        int sLen=s.size();
        int pLen=p.size();
        // if (pLen > 30000) return false; // the trick
        if(pLen==0)
            return sLen==0;
        
        if(p[0]=='*')  
        {
            while(p[0]=='*')
                p=p.substr(1);
            while(sLen>0)
            {
                if(isMatch(s,p))
                    return true;
                sLen--;
                s=s.substr(1);
            }
            return isMatch(s,p);
        }
        else if(p[0]==s[0] || (sLen>0 && p[0]=='?'))
            return isMatch(s.substr(1),p.substr(1));
        
        return false;
    }
 */
    
     bool isMatch(string s, string p) {
        int sLen=s.size();
        int pLen=p.size();
        
        int i=0,j=0,pi=0,pj=pLen;
        
        while(i<sLen)
        {
            if(s[i]==p[j]||p[j]=='?')
            {
                i++;
                j++;
                continue;
            }
            
            if(p[j]=='*'){
                pj=j++;
                pi=i;
                continue;
            }
            
            if(pj<pLen){
                j=pj+1;//j是*后面的字符，i从pi开始逐个匹配
                i=++pi;
                continue;
            }
            return false;
        }
       while(p[j]=='*'){
           j++;
       }
       return j==pLen;
    }
};