题目要求：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

参考网址：http://blog.csdn.net/pickless/article/details/9787227

代码如下：

class Solution {
public:
bool initCheck(const char *s, const char *p) {
int l1 = 0;
int idx = 0, l2 = 0;

    while (s\[l1\] != '\\0') {  
        l1++;  
    }  
    while (p\[idx\] != '\\0') {  
        l2 += p\[idx++\] != '\*' ? 1 : 0;  
    }

    return l1 >= l2;  
}

bool isMatch(const char \*s, const char \*p) {  
    // Start typing your C/C++ solution below  
    // DO NOT write int main() function  
    if (!initCheck(s, p)) {  
        return false;  
    } 

    int l = 0;  
    while (p\[l++\] != '\\0');

    bool prev\[l\], f\[l\];  
    memset(f, false, sizeof(bool) \* l);  
    memset(prev, false, sizeof(bool) \* l);

    bool isFirst = true;  
    for (int i = 0; i < l; i++) {  
        if (p\[i\] == '\*') {  
            f\[i\] = i == 0 || f\[i - 1\];  
        }  
        else if ((p\[i\] == '?' || p\[i\] == \*s) && isFirst) {  
            isFirst = false;  
            f\[i\] = true;  
        }  
        else {  
            break;  
        }  
    }  
    s++;

    while (\*(s - 1) != '\\0') {  
        memcpy(prev, f, l);  
        memset(f, false, sizeof(bool) \* l);

        for (int i = 0; i < l; i++) {  
            if (prev\[i\]) {  
                f\[i\] = f\[i\] || p\[i\] == '\*';  
                f\[i + 1\] = f\[i + 1\] || p\[i + 1\] == '?' || p\[i + 1\] == \*s;  
            }  
            else if (i == 0 || f\[i - 1\]) {  
                f\[i\] = f\[i\] || p\[i\] == '\*';  
            }  
        }  
        s++;  
    }    

    return f\[l - 1\];  
}  

};