問題
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.
制約
0 ≤ number of operations ≤ 10,000
−10^9 ≤ x ≤ 10^9
例
入力：
operations = ["push_back 2", "push_front 1", "push_back 3", "reverse"]
期待される出力：
expected = [3, 2, 1]入力：
std::vector<std::string> operations = {"push_back 2", "push_front 1", "push_back 3", "reverse"};
期待される出力：
std::vector<long long> expected = {3, 2, 1};入力：
string[] operations = new string[] {"push_back 2", "push_front 1", "push_back 3", "reverse"};
期待される出力：
long[] expected = new long[] {3, 2, 1};解説
The list grows to 1 → 2 → 3, and reversing it gives [3, 2, 1].
入力：
operations = ["push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"]
期待される出力：
expected = [5]入力：
std::vector<std::string> operations = {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
期待される出力：
std::vector<long long> expected = {5};入力：
string[] operations = new string[] {"push_back 5", "push_back 7", "push_back 5", "remove 5", "pop_front"};
期待される出力：
long[] expected = new long[] {5};解説
remove takes out only the first 5, leaving 7 → 5; pop_front then removes 7: [5].
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
A list type with cheap front and back inserts (std::list, LinkedList<T>, deque) does most of the work.
ヒント 2
remove deletes only the first match, and pop_front on an empty list does nothing.
ヒント 3
Reversing changes which end is the front for later pushes and pops.
目安の計算量
O(n²) time, O(n) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。