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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 <= 201 <= p.length <= 30scontains only lowercase English letters.pcontains 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;
}
}