LeetCode 044 Wildcard Matching

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];

    }

};