Regular Expression Matching

Description

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.

Code

Top Down DP with memo

• 對於 index i 來說
  • 如果目前字串（s）和 regex（p）相同，比較 index i + 1 之後的部分
    • dp[i][j] = dp[i + 1][j + 1]
  • 如果 p 目前是 . 代表可以 match 任何字母，所以一樣比較 index i + 1 之後的部分
    • dp[i][j] = dp[i + 1][j + 1]
  • 如果 p 在 i + 1 的部分是 star ，一個 star 代表可以 match 0 或是多個 preceding character，例如 a* 代表可以 match 0 或是多個 a，因此分成兩種 case
    • match 0: 比較 i + 2 之後的部分
      • dp[i][j] = dp[i][j + 2]
    • match 1 ~ n：比較 i + 1, i + 2, i + 3 … … 之後的部分
      • dp[i][j] = dp[i + 1][j + 2] || dp[i + 2][j + 2] || dp[i + 3][j + 2] || ......

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        vector<vector<int>> memo(m + 1, vector<int>(n + 1, -1));
        return isMatch(0, 0, s, p, memo);
    }
 
    bool isMatch(int i, int j, string& s, string& p, vector<vector<int>>& memo) {
        int sn = s.length(), pn = p.length();
        if(j == pn) return i == sn;
        if(memo[i][j] != -1) return memo[i][j];
        if(p[j + 1] == '*') {
            if(isMatch(i, j + 2, s, p, memo)) return memo[i][j] = true; // case: *a match zero a
            while(i < sn && (p[j] == s[i] || p[j] == '.')) { // case: *a match one or more a
                if(isMatch(++i, j + 2, s, p, memo)) return memo[i][j] = true;
            }
        } else if(i < sn && (p[j] == '.' || s[i] == p[j])) {
            return memo[i][j] = isMatch(i + 1, j + 1, s, p, memo);
        }
        return memo[i][j] = false;
    }
};

Bottom Up

和 Edit Distance 以及 Distinct Subsequences 類似。

Time Complexity: $O(nm)$, Space Complexity: $O(nm)$

注意 for(int i = m ; i >= 0; i--) 是從 i = m 開始，以 s = "aa", p = "a* 去 dry run 就會知道為何。

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
        dp[m][n] = true;
        for(int i = m ; i >= 0; i--) {
            for(int j = n - 1; j >= 0; j--) {
                if(p[j + 1] == '*') {
                    dp[i][j] = dp[i][j + 2] || i < m && (p[j] == '.' || s[i] == p[j]) && dp[i + 1][j];
                } else {
                    dp[i][j] = i < m && (p[j] == '.' || s[i] == p[j]) && dp[i + 1][j + 1];
                }  
            }
        }
        return dp[0][0];
    }
 
 
};

Source

• Regular Expression Matching - LeetCode