This Solution use 2D DP. beat 90% solutions, very simple.

Here are some conditions to figure out, then the logic can be very straightforward.

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

class Solution {
    public boolean isMatch(String s, String p) {
        if (s == null && p == null) return true;
        int n = s.length(), m = p.length();
        boolean[][] dp = new boolean[n + 1][m + 1];
        dp[0][0] = true;

        for (int j = 1; j <= m; j++){
            if (p.charAt(j - 1) == '*'){
                dp[0][j] = dp[0][j - 1] || (j >= 2 && dp[0][j - 2]);
            }
        }

        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= m; j++){
                if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.'){
                    dp[i][j] = dp[i - 1][j - 1];
                }
                if (p.charAt(j - 1) == '*'){
                    if (s.charAt(i - 1) != p.charAt(j - 2) && p.charAt(j - 2) != '.'){
                        dp[i][j] = dp[i][j - 2];
                    }
                    else{
                        // multiple: dp[i - 1][j]
                        // single: dp[i][j - 1]
                        // none: dp[i][j - 2]
                        dp[i][j] = dp[i - 1][j] || dp[i][j - 1] || dp[i][j - 2];
                    }
                }
            }
        }

        return dp[n][m];
    }
}