LeetCode 10. Regular Expression Matching

Backtracking

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:

• 0 <= s.length <= 20

• 0 <= 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:

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class Solution {
    
    bool isMatch(const string& s, int sidx, const string& p, int pidx) {
        if (pidx >= p.length()) {
            return sidx == s.length();
        }
        if ((pidx + 1) < p.length() && p[pidx + 1] == '*') {
            if (isMatch(s, sidx, p, pidx + 2)) {
                return true;
            }
            if (sidx < s.length() && (p[pidx] == '.' || s[sidx] == p[pidx])) {
                return isMatch(s, sidx + 1, p, pidx);
            }
        } else if (sidx < s.length() && (p[pidx] == '.' || s[sidx] == p[pidx])) {
            return isMatch(s, sidx + 1, p, pidx + 1);
        }
        return false;
    }
    
public:
    bool isMatch(string s, string p) {
        return isMatch(s, 0, p, 0);
    }
};