10、Regular Expression Matching

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

class Solution {public:    bool isMatch(const char *s, const char *p)    {        //此方法是网上的一个比较简单的方法，原理与下面的方法差不多,没通过，不想研究了，以后研究吧        /* if (*p == '\0') return *s == '\0';        if (*(p + 1) != '*')        {            return ((*p == *s) || (*p == '.' && *s != '\0')) && isMatch(s + 1, p + 1);        }        while ((*p == *s) || (*p == '.' && *s == '\0'))        {            if (isMatch(s, p + 2)) return true;            s++;        }        return isMatch(s, p + 2);        */    }    /*bool isMatch(const char *s, const char *p) {        //刚开始题目有些理解错误，把'*'理解为匹配任意并任意多的字符，在此题中是匹配任意多的前缀字符，注意题目意思，个人感觉此方法不是最优的，存在重复计算，更优的解留给以后        int len1 = strlen(s), len2 = strlen(p);        return ismatch(s, p, 0, 0, len1, len2);            }    bool ismatch(const char *s, const char *p, int index1, int index2, int length1, int length2)    {        int i = index1, j = index2;        while (i < length1 && j < length2)        {            //首先我把p[j]分为两种情况，一种是后面带'*', 一种是独立的字母的            if (j + 1 < length2 && p[j+1] == '*')            {                if (ismatch(s, p, i, j + 2, length1, length2) == false)                {                    if (s[i] == p[j] || p[j] == '.')                        ++i;                    else                        return false;                }                else                    return true;            }            else            {                if (s[i] == p[j] || p[j] == '.')                {                    ++i;                    ++j;                }                else                    return false;            }                    }        while (j + 1 < length2 && p[j + 1] == '*')            j += 2;         if (i == length1 && j == length2)            return true;        else            return false;    }*/};