Условие
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.
Ограничения
0 ≤ number of operations ≤ 10,000
−10^9 ≤ x ≤ 10^9
Примеры
Ввод:
operations = ["push_back 2", "push_front 1", "push_back 3", "reverse"]
Ожидаемый вывод:
expected = [3, 2, 1]
Ввод:
std::vector<std::string> operations = {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Ожидаемый вывод:
std::vector<long long> expected = {3, 2, 1};
Ввод:
string[] operations = new string[] {"push_back 2", "push_front 1", "push_back 3", "reverse"};
Ожидаемый вывод:
long[] expected = new long[] {3, 2, 1};
Пояснение
The list grows to 1 → 2 → 3, and reversing it gives [3, 2, 1].
Ввод:
operations = ["push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"]
Ожидаемый вывод:
expected = [5]
Ввод:
std::vector<std::string> operations = {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Ожидаемый вывод:
std::vector<long long> expected = {5};
Ввод:
string[] operations = new string[] {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
Ожидаемый вывод:
long[] expected = new long[] {5};
Пояснение
remove takes out only the first 5, leaving 7 → 5; pop_front then removes 7: [5].
Как подступиться
Сначала попробуйте сами, затем открывайте подсказки по одной.
Подсказка 1
A list type with cheap front and back inserts (std::list, LinkedList<T>, deque) does most of the work.
Подсказка 2
remove deletes only the first match, and pop_front on an empty list does nothing.
Подсказка 3
Reversing changes which end is the front for later pushes and pops.
Ожидаемая сложность
O(n²) time, O(n) extra space
Ссылки
Это задание адаптировано для CodeUtility. Следуйте условию и ограничениям выше.