Problem
A linked list stores values in nodes that point to the next node, so adding or removing at the front is cheap. Simulate one that starts empty.
Process the operations in order:
push_front x: add x at the front.push_back x: add x at the back.pop_front: remove the first value (ignored when the list is empty).remove x: remove the first node whose value is x, if there is one.reverse: reverse the order of the nodes.
Implement linked_list_operations(operations).
Input
operations: The operations.
Output
Return the values in the list from front to back.
Einschränkungen
0 ≤ number of operations ≤ 10,000
−10^9 ≤ x ≤ 10^9
Beispiele
Eingabe:
operations = ["push_back 2", "push_front 1", "push_back 3", "reverse"]
Erwartete Ausgabe:
expected = [3, 2, 1]Eingabe:
std::vector<std::string> operations = {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Erwartete Ausgabe:
std::vector<long long> expected = {3, 2, 1};Eingabe:
string[] operations = new string[] {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Erwartete Ausgabe:
long[] expected = new long[] {3, 2, 1};Erläuterung
The list grows to 1 → 2 → 3, and reversing it gives [3, 2, 1].
Eingabe:
operations = ["push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"]
Erwartete Ausgabe:
expected = [5]Eingabe:
std::vector<std::string> operations = {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Erwartete Ausgabe:
std::vector<long long> expected = {5};Eingabe:
string[] operations = new string[] {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Erwartete Ausgabe:
long[] expected = new long[] {5};Erläuterung
remove takes out only the first 5, leaving 7 → 5; pop_front then removes 7: [5].
So gehst du vor
Versuch es zuerst allein und öffne dann einen Hinweis nach dem anderen.
Hinweis 1
A list type with cheap front and back inserts (std::list, LinkedList<T>, deque) does most of the work.
Hinweis 2
remove deletes only the first match, and pop_front on an empty list does nothing.
Hinweis 3
Reversing changes which end is the front for later pushes and pops.
Erwartete Komplexität
O(n²) time, O(n) extra space
Referenzen
Diese Aufgabe wurde für CodeUtility angepasst. Beachte die oben genannten Vorgaben und Einschränkungen.