Difficulty

Hard

Description

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 = "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

Solution

class Solution {
    public boolean isMatch(String s, String p) {
        if (p.isEmpty())
            return s.isEmpty();
        boolean first_match = (!s.isEmpty() && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.'));  //s與p第一位字元匹配情況
        if (p.length() >= 2 && p.charAt(1) == '*')
            return (isMatch(s, p.substring(2)) || (first_match && isMatch(s.substring(1),p)));    //忽略p中帶*部分或刪去s已匹配部分
        else 
            return first_match && isMatch(s.substring(1), p.substring(1));
    }
    
    
}