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 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 = "cab"
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 = "misisp*."
Output: false

分析：

Java语言：

public static boolean isMatch(String s, String p) {
        if (p.isEmpty()) {
            return s.isEmpty();
        }
        boolean firstMatch = !s.isEmpty() && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.');
        if (p.length() >= 2 && p.charAt(1) == '*') {
            return isMatch(s, p.substring(2)) || (firstMatch && isMatch(s.substring(1), p));
        } else {
            return firstMatch && isMatch(s.substring(1), p.substring(1));
        }
    }