LeetCode-in-All
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
defmodule Solution do
@cache Agent
@spec is_match(s :: String.t, p :: String.t) :: boolean
def is_match(s, p) do
:ets.new(:cache, [:named_table, :set, {:read_concurrency, true}])
result = is_match(s, p, 0, 0)
:ets.delete(:cache)
result
end
defp is_match(s, p, i, j) do
case :ets.lookup(:cache, {i, j}) do
[{_, result}] -> result
[] ->
result =
if j == String.length(p) do
i == String.length(s)
else
first_match = i < String.length(s) && (String.at(s, i) == String.at(p, j) || String.at(p, j) == ".")
if (j + 1) < String.length(p) && String.at(p, j + 1) == "*" do
(first_match && is_match(s, p, i + 1, j)) || is_match(s, p, i, j + 2)
else
first_match && is_match(s, p, i + 1, j + 1)
end
end
:ets.insert(:cache, { {i, j}, result})
result
end
end
end