### LeetCode - Regular Expression Matching

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

1. 当*表示0时，我们只需要得到dp[i][j-2]的值即可。

2. 当*表示1时，我们需要判断s.charAt(i - 1), p.charAt(j-2)是否相等，若相等的话，则dp[i][j]其值为dp[i-1][j-2]。

3. 当*表示>1时，我们需要判断s.charAt(i - 1), p.charAt(j-2)是否相等，若相等的话，则dp[i][j]其值为dp[i-1][j]。

`package dp;public class RegularExpressionMatching {    public boolean isMatch(String s, String p) {        int m = s.length();        int n = p.length();        boolean[][] dp = new boolean[m + 1][n + 1];                dp[0][0] = true;        for (int i = 1; i <= n; ++i)            dp[0][i] = p.charAt(i - 1) == '*' ? dp[0][i - 2] : false;        for (int i = 1; i <= m; ++i) {            for (int j = 1; j <= n; ++j) {                if (p.charAt(j - 1) == '*') {                    dp[i][j] = dp[i][j-2] ||                             (isCharMatch(s.charAt(i - 1), p.charAt(j-2)) && dp[i-1][j-2]) ||                             (isCharMatch(s.charAt(i - 1), p.charAt(j-2)) && dp[i-1][j]);                } else {                    dp[i][j] = isCharMatch(s.charAt(i - 1), p.charAt(j - 1)) && dp[i-1][j-1];                }            }        }                return dp[m][n];    }        public boolean isCharMatch(char a, char b) {        if (b == '.') return true;        return a == b;    }        public static void main(String[] args) {        // TODO Auto-generated method stub        RegularExpressionMatching r = new RegularExpressionMatching();        System.out.println(r.isMatch("aaa", "a*"));    }}`