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.

 动态规划，无敌存在

class Solution {public:    bool isMatch(const char *s, const char *p) {        int i, j;        int m = strlen(s);        int n = strlen(p);        /**         * b[i + 1][j + 1]: if s[0..i] matches p[0..j]         * if p[j] != '*'         * b[i + 1][j + 1] = b[i][j] && s[i] == p[j]         * if p[j] == '*', denote p[j - 1] with x,         * then b[i + 1][j + 1] is true iff any of the following is true         * 1) "x*" repeats 0 time and matches empty: b[i + 1][j -1]         * 2) "x*" repeats 1 time and matches x: b[i + 1][j]         * 3) "x*" repeats >= 2 times and matches "x*x": s[i] == x && b[i][j + 1]         * '.' matches any single character         */        bool b[m + 1][n + 1];        b[0][0] = true;        for (i = 0; i < m; i++) {            b[i + 1][0] = false;        }        // p[0..j - 2, j - 1, j] matches empty iff p[j] is '*' and p[0..j - 2] matches empty        for (j = 0; j < n; j++) {            b[0][j + 1] = j > 0 && '*' == p[j] && b[0][j - 1];        }        for (i = 0; i < m; i++) {            for (j = 0; j < n; j++) {                if (p[j] != '*') {                    b[i + 1][j + 1] = b[i][j] && ('.' == p[j] || s[i] == p[j]);                } else {                    b[i + 1][j + 1] = b[i + 1][j - 1] && j > 0 || b[i + 1][j] ||                                b[i][j + 1] && j > 0 && ('.' == p[j - 1] || s[i] == p[j - 1]);                }            }        }        return b[m][n];    }};