Skip to content

Commit

Permalink
regular-expression-matching
Browse files Browse the repository at this point in the history
  • Loading branch information
aolenevme authored Mar 1, 2023
1 parent 80dfd90 commit ac863f4
Showing 1 changed file with 53 additions and 0 deletions.
53 changes: 53 additions & 0 deletions string/regular-expression-matching.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,53 @@
/**
10. Regular Expression Matching
Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:
'.' 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 (.)".
Constraints:
1 <= s.length <= 20
1 <= p.length <= 20
s contains only lowercase English letters.
p contains only lowercase English letters, '.', and '*'.
It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.
*/

/**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
var isMatch = function (s, p) {
if (p.length === 0) {
return s.length === 0;
}

const isFirstMatched = s.length > 0 && (s[0] === p[0] || p[0] === ".");

if (p.length >= 2 && p[1] === "*") {
return (
isMatch(s, p.substring(2)) ||
(isFirstMatched && isMatch(s.substring(1), p))
);
} else {
return isFirstMatched && isMatch(s.substring(1), p.substring(1));
}
};

0 comments on commit ac863f4

Please sign in to comment.