------
## 最终结论
```python
def is_match(s: str, p: str) -> bool:
    # Create a DP table with dimensions (len(s)+1) x (len(p)+1)
    dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
    
    # Base case initialization
    dp[0][0] = True
    
    # Fill the first row where s is empty
    for j in range(1, len(p) + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 2]
    
    # Fill the rest of the DP table
    for i in range(1, len(s) + 1):
        for j in range(1, len(p) + 1):
            if p[j - 1] == '.' or p[j - 1] == s[i - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            elif p[j - 1] == '*':
                dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] if (p[j - 2] == s[i - 1] or p[j - 2] == '.') else False)
    
    # Return the result of the matching
    return dp[len(s)][len(p)]

# Example usage:
s = "aab"
p = "c*a*b"
print(is_match(s, p))  # Output: True

s = "mississippi"
p = "mis*is*p*."
print(is_match(s, p))  # Output: False
```