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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
object Solution {
private var cache: Array[Array[Option[Boolean]]] = Array.ofDim[Option[Boolean]](0, 0)
def isMatch(s: String, p: String): Boolean = {
cache = Array.ofDim[Option[Boolean]](s.length + 1, p.length + 1)
isMatch(s, p, 0, 0)
}
private def isMatch(s: String, p: String, i: Int, j: Int): Boolean = {
if (j == p.length) {
return i == s.length
}
var result: Boolean = false
if (cache(i).isDefinedAt(j) && cache(i)(j) != null) {
return cache(i)(j).get
}
val firstMatch = i < s.length && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.')
if (j + 1 < p.length && p.charAt(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) = Some(result)
result
}
}