Regular Expression Matching

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转自：http://blog.csdn.net/doc_sgl/article/details/12719761

Implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.'*' Matches zero or more of the preceding element.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", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

注意：这里的a*表示a可以重复0次或者多次，不是a和*分开的。

It seems that some readers are confused about why the regex pattern ".*" matches the string"ab". ".*" means repeat the preceding element 0 or more times. Here, the "preceding" element is the dot character in the pattern, which can match any characters. Therefore, the regex pattern".*" allows the dot to be repeated any number of times, which matches any string (even an empty string). Think carefully how you would do matching of '*'.Please note that '*' in regular expression is different from wildcard matching, as we match the previous character 0 or more times. But, how many times? If you are stuck,recursion is your friend.

#include <iostream>#include<cmath>#include<cstring>using namespace std;class Solution {public:    bool isMatch(const char *s, const char *p)    {        //end flag        if(*p == '\0') return *s == '\0';        if(*(p+1) != '*')        {            if(*s != '\0' && (*s == *p || *p == '.') )                return isMatch(s+1,p+1);            else                return false;        }        else        {            //对于*重复次数的迭代处理            while(*s != '\0' && (*s == *p || *p == '.'))            {                if(isMatch(s,p+2)) return true;                s++;            }            return isMatch(s,p+2);        }    }};int main(){    const char a[10] = "aaaabb";    const char b[10] = "a*bb";    Solution so;    cout << so.isMatch(a,b) << endl;    return 0;}

之前一直没理解，发现对if和return那两句有理解偏差。

if并没有改变p的指针，可以理解为assert。对于return，实际值指针已经改变。

#include <iostream>#include<cmath>#include<cstring>using namespace std;class Solution {public:    bool isMatchTest(const char *s)    {        if(*s == '\0') return true;        if(*(s+1) != '\0')            cout<<*s<<endl;        return isMatchTest(s+2);    }};int main(){    const char a[10] = "abcdef";    Solution so;    so.isMatchTest(a);    //输出ace    return 0;}

另外加了一段正常通配符可用的程序。？->一个字符 *->任意字符

#include <iostream>#include<cmath>#include<cstring>using namespace std;class Solution {public:    bool isMatch(const char *s, const char *p)    {        //end flag        if(*p == '\0') return *s == '\0';        if(*p != '*')        {            if(*s != '\0' && (*s == *p || *p == '?') )                return isMatch(s+1,p+1);            else                return false;        }        else        {            //对于*重复次数的迭代处理            /* if there is *, then there are two possibilities       a) We consider current character of second string       b) We ignore current character of second string.*/            return isMatch(s,p+1)||isMatch(s+1,p);        }    }};int main(){    const char a[10] = "txt.xls";    const char b[10] = "t?t.*";    Solution so;    cout << so.isMatch(a,b) << endl;    return 0;}

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