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# Wildcard Matching

Given an input string (`s`) and a pattern (`p`), implement wildcard pattern matching with support for `'?'` and `'*'`.

```
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
```

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 = "*"
Output: true
Explanation: '*' matches any sequence.
```

**Example 3:**

```
Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.
```

**Example 4:**

```
Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
```

**Example 5:**

```
Input:
s = "acdcb"
p = "a*c?b"
Output: false
```

```java
class Solution {
    public boolean isMatch(String s, String p) {
        if (s == null || p == null)
            return false;
        int sLen = s.length();
        int pLen = p.length();
        boolean[][] dp = new boolean[sLen + 1][pLen + 1];

        // Base cases:
        dp[0][0] = true;
        for (int i = 1; i <= sLen; i++)
            dp[i][0] = false;

        for (int j = 1; j <= pLen; j++) 
            if (p.charAt(j - 1) == '*') 
                dp[0][j] = dp[0][j - 1];

        // Recursion:
        for (int i = 1; i <= sLen; i++) {
            for (int j = 1; j <= pLen; j++) {
                if ((s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') && dp[i - 1][j - 1])
                    dp[i][j] = true;
                else if (p.charAt(j - 1) == '*' && (dp[i - 1][j] || dp[i][j - 1]))
                    dp[i][j] = true;
            }   // dp[i][j-1] is the case when * -> ""
                // dp[i-1][j] is the case when * copies subsequence of s
        }
        return dp[sLen][pLen];
    }
}
```