Problema
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.
Restricciones
0 ≤ number of operations ≤ 10,000
−10^9 ≤ x ≤ 10^9
Ejemplos
Entrada:
operations = ["push_back 2", "push_front 1", "push_back 3", "reverse"]
Salida esperada:
expected = [3, 2, 1]Entrada:
std::vector<std::string> operations = {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Salida esperada:
std::vector<long long> expected = {3, 2, 1};Entrada:
string[] operations = new string[] {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Salida esperada:
long[] expected = new long[] {3, 2, 1};Explicación
The list grows to 1 → 2 → 3, and reversing it gives [3, 2, 1].
Entrada:
operations = ["push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"]
Salida esperada:
expected = [5]Entrada:
std::vector<std::string> operations = {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Salida esperada:
std::vector<long long> expected = {5};Entrada:
string[] operations = new string[] {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Salida esperada:
long[] expected = new long[] {5};Explicación
remove takes out only the first 5, leaving 7 → 5; pop_front then removes 7: [5].
Cómo abordarlo
Inténtalo primero por tu cuenta y luego abre las pistas una a una.
Pista 1
A list type with cheap front and back inserts (std::list, LinkedList<T>, deque) does most of the work.
Pista 2
remove deletes only the first match, and pop_front on an empty list does nothing.
Pista 3
Reversing changes which end is the front for later pushes and pops.
Complejidad esperada
O(n²) time, O(n) extra space
Referencias
Este reto está adaptado para CodeUtility. Sigue el enunciado y las restricciones indicadas arriba.