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

Tips:

1. If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1];
2. If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];

3. If p.charAt(j) == '*':

   here are two sub conditions:

     1.if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //in this case, a* only counts as empty
     2.if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
                          dp[i][j] = dp[i-1][j]    //in this case, a* counts as multiple a 
                       or dp[i][j] = dp[i][j-1]   // in this case, a* counts as single a
                       or dp[i][j] = dp[i][j-2]   // in this case, a* counts as empty
   

Code:

 public boolean isMatch(String s, String p) {

    if (s == null || p == null) {
        return false;
    }
    boolean[][] dp = new boolean[s.length()+1][p.length()+1];
    dp[0][0] = true;
    for (int i = 0; i < p.length(); i++) {
        if (p.charAt(i) == '*' && dp[0][i-1]) {
            dp[0][i+1] = true;
        }
    }
    for (int i = 0 ; i < s.length(); i++) {
        for (int j = 0; j < p.length(); j++) {
            if (p.charAt(j) == '.') {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == s.charAt(i)) {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == '*') {
                if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
                    dp[i+1][j+1] = dp[i+1][j-1];
                } else {
                    dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
                }
            }
        }
    }
    return dp[s.length()][p.length()];
}