10 Regular Expression Matching *****DP hard !!!!IMP(FB)

Given an input string (s) and a pattern (p), 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).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.
Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

class Solution {
    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) == s.charAt(i)) {
                    dp[i + 1][j + 1] = dp[i][j];
                }
                if (p.charAt(j) == '.') {
                    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()];
    }
}