LeetCode Algorithm 0010 - Regular Expression Matching (Hard)

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LeetCode Algorithm 0010 - Regular Expression Matching (Hard)

Problem Link: https://leetcode.com/problems/regular-expression-matching/description/

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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:

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Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

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Solution C++

#pragma once

#include "pch.h"

// Problem: https://leetcode.com/problems/palindrome-number/description/

namespace P10RegularExpressionMatching
{
    class Solution
    {
        private:
        static const char k_Star = '*';
        static const char k_Dot = '.';

        public:
        bool isMatch(string s, string p)
        {
            // all empty
            if (p.empty())
            {
                return s.empty();
            }

            bool firstMatch = !s.empty() && (p.at(0) == k_Dot || p.at(0) == s.at(0));

            if (p.size() > 1 && p.at(1) == k_Star)
            {
                return isMatch(s, p.substr(2)) || firstMatch && isMatch(s.substr(1), p);
            }
            else
            {
                return firstMatch && isMatch(s.substr(1), p.substr(1));
            }
        }
    };
}

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