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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

 

解法一：

这题想法参考了，

类似于一种有限状态机的做法。

主要思想如下：

由于*匹配多少个字符是不一定的，于是首先假设*不匹配任何字符。

当后续匹配过程中出错，采用回溯的方法，假设*匹配0个字符、1个字符、2个字符……i个字符，然后继续匹配。

因此s需要有一个spos指针，记录当p中的*匹配i个字符后，s的重新开始位置。

p也需要一个starpos指针指向*的位置，当回溯过后重新开始时，从starpos的下一位开始。

class Solution {public:    bool isMatch(const char *s, const char *p) {        char* starpos = NULL;        char* spos = NULL;                while(true)        {            if(*s == 0 && *p == 0)            //match all                return true;            else if(*s == 0 && *p != 0)            {
    //successive *p must be all '*'                while(*p != 0)                {                    if(*p != '*')                        return false;                    p ++;                }                return true;            }            else if(*s != 0 && *p == 0)            {                if(starpos != NULL)                {
    //maybe '*' matches too few chars in s                    p = starpos + 1;                    s = spos + 1;                    spos ++;    //let '*' matches one more chars in s                }                else                //*s is too long                    return false;            }            else            {                if(*p == '?' || *s == *p)                {                    s ++;                    p ++;                }                else if(*p == '*')                {                    starpos = (char*)p;                    spos = (char*)s;                    p ++;                       //start successive matching from "'*' matches non"                }                //currently not match                else if(starpos != NULL)                {
    //maybe '*' matches too few chars in s                    p = starpos + 1;                    s = spos + 1;                    spos ++;    //let '*' matches one more chars in s                }                else                //not match                    return false;            }        }    }};

 

解法二：

模仿的递归做法，小数据集上能过。

当*数目太多时会超时。

class Solution {public:    bool isMatch(const char *s, const char *p) {        if(*p == 0)            return (*s == 0);        if(*p == '*')        {            //case1: *p matches 0 chars in s            if(isMatch(s, p+1))                return true;                            //case2: try all possible numbers            while(*s != 0)            {                s ++;                if(isMatch(s, p+1))                    return true;            }            return false;        }        else        {            if((*s==*p) || (*p=='?'))                return isMatch(s+1, p+1);            else                return false;        }    }};

以下是我的测试代码，小数据上全部通过：

int main(){    Solution s;    cout << s.isMatch("aa","a") << endl;    cout << s.isMatch("aa","aa") << endl;    cout << s.isMatch("aaa","aa") << endl;    cout << s.isMatch("aa","*") << endl;    cout << s.isMatch("aa","a*") << endl;    cout << s.isMatch("ab","?*") << endl;    cout << s.isMatch("aab","c*a*b") << endl;    return 0;}