[Leetcode] 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") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("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];
    }
};