LeetCode in Net

10. Regular Expression Matching

Hard

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 (.)”.

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

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• 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.

Solution

public class Solution {
    private bool?[,] cache;

    public bool IsMatch(string s, string p) {
        cache = new bool?[s.Length + 1, p.Length + 1];
        return IsMatch(s, p, 0, 0);
    }

    private bool IsMatch(string s, string p, int i, int j) {
        if (j == p.Length) {
            return i == s.Length;
        }
        bool result;
        if (cache[i, j] != null) {
            return cache[i, j].Value;
        }
        bool firstMatch = i < s.Length && (s[i] == p[j] || p[j] == '.');
        if ((j + 1) < p.Length && p[j + 1] == '*') {
            result = (firstMatch && IsMatch(s, p, i + 1, j)) || IsMatch(s, p, i, j + 2);
        } else {
            result = firstMatch && IsMatch(s, p, i + 1, j + 1);
        }
        cache[i, j] = result;
        return result;
    }
}

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