Regular Expression Matching

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 preceding 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) {
        int n = s.length(), m = p.length();
        boolean [][]f = new boolean[n + 2][m + 2];
        f[0][0] = true;
        for (int j = 1; j <= m; ++j) {
            for (int i = 0; i <= n; ++i) {
                if (p.charAt(j - 1) >= 'a' && p.charAt(j - 1) <= 'z') {
                    if (i >= 1 && 
                        p.charAt(j - 1) == s.charAt(i - 1) && f[i - 1][j - 1]) {
                        f[i][j] = true;
                    } 
                } else if (p.charAt(j - 1) == '.') {
                    if (i >= 1 && f[i - 1][j - 1]) {
                        f[i][j] = true;
                    }
                } else if (p.charAt(j - 1) == '*') {
                    if (f[i][j - 1]) f[i][j] = true;
                    if (i >= 1 
                        && (p.charAt(j - 2) == '.' 
                            ||p.charAt(j - 2) == s.charAt(i - 1) )
                        && f[i - 1][j - 1]) 
                        f[i][j] = true;
                    if (f[i][j - 2]) f[i][j] = true;
                    if (i >= 1 && (p.charAt(j - 2) == '.' ||
                        p.charAt(j - 2) == s.charAt(i - 1) )
                        && f[i - 1][j]) 
                        f[i][j] = true;
                }
            }
        }
        return f[n][m];
    }
}