문제
In C++, RAII ties a resource to an object's lifetime: the resource is released automatically when the object goes out of scope. Objects in the same scope are destroyed in reverse order of creation.
Process the events in order:
{: enter a new inner scope.}: leave the current scope; every resource acquired in it is released, newest first.acquire <name>: create a resource in the current scope.
When the events run out, the remaining open scopes end from the innermost to the outermost (the program's outer scope included), releasing their resources the same way.
Implement raii_release_order(events).
Input
events: The events. Braces are balanced or left open, never closed without being opened.
Output
Return the resource names in the order they are released.
제약 조건
0 ≤ number of events ≤ 100,000
Names have 1 to 20 characters without spaces
예시
입력:
events = ["acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"]
예상 출력:
expected = ["lock", "file", "log", "db"]
입력:
std::vector<std::string> events = {"acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"};
예상 출력:
std::vector<std::string> expected = {"lock", "file", "log", "db"};
입력:
string[] events = new string[] {"acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"};
예상 출력:
string[] expected = new string[] {"lock", "file", "log", "db"};
설명
Leaving the inner scope releases lock then file (newest first). At the end the outer scope releases log then db: ["lock", "file", "log", "db"].
입력:
events = ["{", "acquire a", "{", "acquire b"]
예상 출력:
expected = ["b", "a"]
입력:
std::vector<std::string> events = {"{", "acquire a", "{", "acquire b"};
예상 출력:
std::vector<std::string> expected = {"b", "a"};
입력:
string[] events = new string[] {"{", "acquire a", "{", "acquire b"};
예상 출력:
string[] expected = new string[] {"b", "a"};
설명
Two scopes are still open at the end: the innermost releases b, then the next releases a: ["b", "a"].
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
Keep a stack of scopes, each holding its resources in order.
힌트 2
Closing a scope releases its resources from newest to oldest.
힌트 3
At the end, close every scope that is still open, innermost first.
예상 복잡도
O(number of events) time, O(number of events) extra space
참고 자료
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