問題
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.
制約
0 ≤ number of events ≤ 100,000
Names have 1 to 20 characters without spaces
例
入力:
events = ["acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"]
期待される出力:
expected = ["lock", "file", "log", "db"]
入力:
std::vector<std::string> events = {"acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"};
期待される出力:
std::vector<std::string> expected = {"lock", "file", "log", "db"};
入力:
string[] events = new string[] {"acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"};
期待される出力:
string[] expected = new string[] {"lock", "file", "log", "db"};
解説
Leaving the inner scope releases lock then file (newest first). At the end the outer scope releases log then db: ["lock", "file", "log", "db"].
入力:
events = ["{", "acquire a", "{", "acquire b"]
期待される出力:
expected = ["b", "a"]
入力:
std::vector<std::string> events = {"{", "acquire a", "{", "acquire b"};
期待される出力:
std::vector<std::string> expected = {"b", "a"};
入力:
string[] events = new string[] {"{", "acquire a", "{", "acquire b"};
期待される出力:
string[] expected = new string[] {"b", "a"};
解説
Two scopes are still open at the end: the innermost releases b, then the next releases a: ["b", "a"].
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Keep a stack of scopes, each holding its resources in order.
ヒント 2
Closing a scope releases its resources from newest to oldest.
ヒント 3
At the end, close every scope that is still open, innermost first.
目安の計算量
O(number of events) time, O(number of events) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。