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 theentireinput 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

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

Here is the solution

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) == '.') {
                    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()];
    }
}