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

动态规划请参考：动态规划解法

递归方法：（TLE）

class Solution {public:    bool isMatch(const char *s, const char *p) {        if( 0 == *p) return 0 == *s;        if(*p == '*'){            while(*(p+1) == '*')p++;            while((*s) != 0){                if(isMatch(s,p+1))return true;                s++;            }            return isMatch(s,p+1);        }        else{            if(((*s) != 0) && ((*s) == (*p) || (*p) == '?')){                return isMatch(s+1,p+1);            }            return false;        }    }};

水平有限，调代码调的比较久，边界问题总是弄不清楚，要死的节奏。
这里当前面和后面均有*号的时候，本来想用kmp,可是KMP也不是很熟，而且这里还有？号。
总体思想是：分析 p 串，把其中每个单词均解析出来，然后用贪心的方法进行匹配，匹配后要保存匹配到 s 中的位置。
总体还是过了，232ms，前面的递归方法TLE。
贪心方法：

class Solution {public:    bool isMatch(const char *s, const char *p) {        //greedy        bool pre,next;        const char *start,*end,*mid;        const char *ss,*se,*sm;        start = p;        end = p;        ss = s;        while(true){            if(*start == '*')pre = true;//查看前面有没有*else pre = false;            //查看新的没有*的word            while(*start != '\0' && *start == '*')start++;            end = start;                        while(*end != '\0' && *end != '*')end++;            if(*end == '*')next = true;else next = false;end--;                        if(*ss == '\0' && pre && *start == '\0')return true;//p和s均完全匹配            if(*ss != '\0' && pre && *start == '\0')return true;            if(*ss != '\0' && (*start == '\0' && !pre))return false;//s没能完全匹配            if(*ss == '\0' && *start != '\0')return false;//p的字母没能匹配完全                        //处理greedy匹配问题            if(next == true){//后面有*可供匹配                if(pre == true){//前面有供做匹配的sm = ss,mid = start;//判断s中剩余的还够匹配不够                    while(mid != end && *sm != '\0'){                        mid++,sm++;                    }                    if(*sm == '\0')return false;mid = start;                    while(*mid != '*' && *ss != '\0'){                        for(sm = ss,mid = start;*sm != '\0' && *mid != '*';sm++,mid++){                            if(*mid == '?')continue;                            if(*mid != *sm)break;                        }                        if(*mid == '*')ss = sm;                        else ss++;                    }//while                    if(*mid != '*')return false;ss = sm;start = mid;                }                else{//前面没有供匹配的                    for(sm = ss,mid = start;*sm != '\0' && mid != end+1;sm++,mid++){                        if(*mid == '?')continue;                        if(*mid != *sm)return false;                    }                    if(mid != end+1)return false;                    ss = sm;//匹配成功start = end+1;                }            }            else{//最后必须匹配的一串，因为后面没有*可供匹配最后的字母if(pre){se = ss;while(*se != '\0')se++;se--;for(sm = se,mid = end;mid != start && sm != ss;mid--,sm--){if(*mid == '?')continue;if(*mid != *sm)return false;}if((sm == ss && mid != start) || (*mid != *sm && *mid != '?'))return false;//如果s不够匹配或者p开头不能匹配return true;}else{for(sm = ss,mid = start;*sm != '\0' && *mid !='\0';sm++,mid++){if(*mid == '?')continue;if(*mid != *sm)return false;}if(*sm != '\0' || *mid != '\0')return false;return true;}                            }        }    }};