0010.regular-expression-matching

0010. Regular Expression Matching

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Q

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 = "mis*is*p*." Output: false

A

class Solution {
    func isMatch(_ s: String, _ p: String) -> Bool {
        if p.isEmpty {
            return s.isEmpty
        }
        let sArr = Array(s), pArr = Array(p)
        let isFirstMatch : Bool = (!s.isEmpty) && (pArr[0] == sArr[0] || pArr[0] == ".")
        if (pArr.count >= 2 && pArr[1] == "*") {
            return isFirstMatch && isMatch(String(s.suffix(s.count - 1)), p) || isMatch(s, String(p.suffix(p.count - 2)))
        } else {
            return isFirstMatch && isMatch(String(s.suffix(s.count - 1)), String(p.suffix(p.count - 1)))
        }
    }
}