โจทย์
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.
เงื่อนไข
0 ≤ number of events ≤ 100,000
Names have 1 to 20 characters without spaces
ตัวอย่าง
อินพุต:
events = ["acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"]
ผลลัพธ์ที่คาดหวัง:
expected = ["lock", "file", "log", "db"]
อินพุต:
std::vector<std::string> events = {"acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"};
ผลลัพธ์ที่คาดหวัง:
std::vector<std::string> expected = {"lock", "file", "log", "db"};
อินพุต:
string[] events = new string[] {"acquire db", "{", "acquire file", "acquire lock", "}", "acquire log"};
ผลลัพธ์ที่คาดหวัง:
string[] expected = new string[] {"lock", "file", "log", "db"};
คำอธิบาย
Leaving the inner scope releases lock then file (newest first). At the end the outer scope releases log then db: ["lock", "file", "log", "db"].
อินพุต:
events = ["{", "acquire a", "{", "acquire b"]
ผลลัพธ์ที่คาดหวัง:
expected = ["b", "a"]
อินพุต:
std::vector<std::string> events = {"{", "acquire a", "{", "acquire b"};
ผลลัพธ์ที่คาดหวัง:
std::vector<std::string> expected = {"b", "a"};
อินพุต:
string[] events = new string[] {"{", "acquire a", "{", "acquire b"};
ผลลัพธ์ที่คาดหวัง:
string[] expected = new string[] {"b", "a"};
คำอธิบาย
Two scopes are still open at the end: the innermost releases b, then the next releases a: ["b", "a"].
แนวทางการแก้
ลองทำเองก่อน แล้วค่อยเปิดคำใบ้ทีละข้อ
คำใบ้ 1
Keep a stack of scopes, each holding its resources in order.
คำใบ้ 2
Closing a scope releases its resources from newest to oldest.
คำใบ้ 3
At the end, close every scope that is still open, innermost first.
ความซับซ้อนที่คาดหวัง
O(number of events) time, O(number of events) extra space
แหล่งอ้างอิง
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