10. Regular Expression Matching

📋 Đề Bài

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 <= 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.

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)Matrix (Ma trận)String (Chuỗi)

⏱️ Thời gian O(n×m)

💾 Không gian O(n×m)

💻 Lời Giải

C++ 0010-regular-expression-matching.cpp

class Solution {
public:
    bool isMatch(string s, string p) {
        int n = (int)s.size();
        int m = (int)p.size();
        
        bool dp[n + 1][m + 1];
        memset(dp, false, sizeof(dp));
        dp[0][0] = true;
        
        for (int j = 2; j <= m; ++j) {
            dp[0][j] = dp[0][j - 2] and p[j - 1] == '*';
        }
        
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (s[i - 1] == p[j - 1] or p[j - 1] == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                }
                else if (p[j - 1] == '*') {
                    dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.'));
                }
            }
        }
        
        return dp[n][m];
    }
};