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

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

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.

//完全不会。。。。。。//http://blog.csdn.net/doc_sgl/article/details/12719761//大概明白一些了class Solution {public:    bool isMatch(string s, string p) {        /**         * f[i][j]: if s[0..i-1] matches p[0..j-1]         * if p[j - 1] != '*'         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]         * if p[j - 1] == '*', denote p[j - 2] with x         *      f[i][j] is true iff any of the following is true         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]         * '.' matches any single character         */        int m = s.size();        int n = p.size();        vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));        f[0][0] = true;//两个空串一定是匹配的                //空串与任意一个非空串都不匹配        for (int i = 1; i <= m; i++)            f[i][0] = false;                //要让任意的串与空串匹配，则要求当前的字符为*，切p[0,...,j-3]与空串匹配            // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty        for (int j = 1; j <= n; j++)            f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];        for (int i = 1; i <= m; i++)            for (int j = 1; j <= n; j++)                //当前的字符不是*，                if (p[j - 1] != '*')                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);                else                    // p[0] cannot be '*' so no need to check "j > 1" here                    f[i][j] = (f[i][j - 2]) || ((s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j]);        return f[m][n];    }};