Problème
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.
Contraintes
0 ≤ number of operations ≤ 10,000
−10^9 ≤ x ≤ 10^9
Exemples
Entrée :
operations = ["push_back 2", "push_front 1", "push_back 3", "reverse"]
Sortie attendue :
expected = [3, 2, 1]Entrée :
std::vector<std::string> operations = {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Sortie attendue :
std::vector<long long> expected = {3, 2, 1};Entrée :
string[] operations = new string[] {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Sortie attendue :
long[] expected = new long[] {3, 2, 1};Explication
The list grows to 1 → 2 → 3, and reversing it gives [3, 2, 1].
Entrée :
operations = ["push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"]
Sortie attendue :
expected = [5]Entrée :
std::vector<std::string> operations = {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Sortie attendue :
std::vector<long long> expected = {5};Entrée :
string[] operations = new string[] {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Sortie attendue :
long[] expected = new long[] {5};Explication
remove takes out only the first 5, leaving 7 → 5; pop_front then removes 7: [5].
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
A list type with cheap front and back inserts (std::list, LinkedList<T>, deque) does most of the work.
Indice 2
remove deletes only the first match, and pop_front on an empty list does nothing.
Indice 3
Reversing changes which end is the front for later pushes and pops.
Complexité attendue
O(n²) time, O(n) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.