10. Regular Expression Matching

package LeetCode_10

/**
 * 10. Regular Expression Matching
 * https://leetcode.com/problems/regular-expression-matching/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).

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 preceding 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 (.)".

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 *.
 * */
class Solution {
    /*
    * solution: recursion, Time complexity:O(n*n), Space complexity:O(n*n)
    * n is min(s.length, p.length)
    * */
    fun isMatch(s: String, p: String): Boolean {
        if (p == null || p.isEmpty()) {
            return s.isEmpty()
        }
        if (s == p) {
            return true
        }
        if (p.length >= 2 && p[1] == '*') {
            //if p's second character is *, so p can match any number of character before *
            if (isMatch(s, p.substring(2))) {
                //check remaining character
                return true
            }
            //otherwise, check first character match or not
            if (s.isNotEmpty() && (s[0] == p[0] || p[0] == '.')) {
                return isMatch(s.substring(1), p)
            }
        } else if (s.isNotEmpty() && (s[0] == p[0] || p[0] == '.')) {
            //need check character one by one
            return isMatch(s.substring(1), p.substring(1))
        }
        return false
    }
}