44. 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

　　

 代码来自：https://leetcode.com/problems/wildcard-matching/discuss/

几种情况：

一：遇到“*” 就记录此时s及p的位置istar及jstar，s向后移，p保持“*”。

二：当s[i]=p[j]或者p[j] = "?"时，就同时向后移，（不在if，也不在else的if中出现，就默认自动同时向后移）

三：遇到s[i] != p[j],而且p也不等于“？”，此时发生不匹配，那就判断p中是否有已出现过“*”，若p中出现过“*”，p回到"*"的位置jstar，s回到istar+1.

class Solution {
public:
    bool isMatch(string s, string p) {
        int  slen = s.size(), plen = p.size(), i, j, iStar=-1, jStar=-1;

        for(i=0,j=0 ; i<slen; ++i, ++j)
        {
            if(p[j]=='*')  //meet a new '*', update traceback i/j info
            { 
                iStar = i;
                jStar = j;
                --i;
            }
            else
            { 
                if(p[j]!=s[i] && p[j]!='?')
                {  // mismatch happens
                    if(iStar >=0)  // met a '*' before, then do traceback
                    { 
                        i = iStar++;
                        j = jStar;
                    }
                    else return false; // otherwise fail
                }
            }
        }
        while(p[j]=='*') ++j;
        return j==plen;
    }
};