{ "data": { "question": { "questionId": "1767", "questionFrontendId": "1670", "categoryTitle": "Algorithms", "boundTopicId": 499831, "title": "Design Front Middle Back Queue", "titleSlug": "design-front-middle-back-queue", "content": "
Design a queue that supports push
and pop
operations in the front, middle, and back.
Implement the FrontMiddleBack
class:
FrontMiddleBack()
Initializes the queue.void pushFront(int val)
Adds val
to the front of the queue.void pushMiddle(int val)
Adds val
to the middle of the queue.void pushBack(int val)
Adds val
to the back of the queue.int popFront()
Removes the front element of the queue and returns it. If the queue is empty, return -1
.int popMiddle()
Removes the middle element of the queue and returns it. If the queue is empty, return -1
.int popBack()
Removes the back element of the queue and returns it. If the queue is empty, return -1
.Notice that when there are two middle position choices, the operation is performed on the frontmost middle position choice. For example:
\n\n6
into the middle of [1, 2, 3, 4, 5]
results in [1, 2, 6, 3, 4, 5]
.[1, 2, 3, 4, 5, 6]
returns 3
and results in [1, 2, 4, 5, 6]
.\n
Example 1:
\n\n\nInput:\n["FrontMiddleBackQueue", "pushFront", "pushBack", "pushMiddle", "pushMiddle", "popFront", "popMiddle", "popMiddle", "popBack", "popFront"]\n[[], [1], [2], [3], [4], [], [], [], [], []]\nOutput:\n[null, null, null, null, null, 1, 3, 4, 2, -1]\n\nExplanation:\nFrontMiddleBackQueue q = new FrontMiddleBackQueue();\nq.pushFront(1); // [1]\nq.pushBack(2); // [1, 2]\nq.pushMiddle(3); // [1, 3, 2]\nq.pushMiddle(4); // [1, 4, 3, 2]\nq.popFront(); // return 1 -> [4, 3, 2]\nq.popMiddle(); // return 3 -> [4, 2]\nq.popMiddle(); // return 4 -> [2]\nq.popBack(); // return 2 -> []\nq.popFront(); // return -1 -> [] (The queue is empty)\n\n\n
\n
Constraints:
\n\n1 <= val <= 109
1000
calls will be made to pushFront
, pushMiddle
, pushBack
, popFront
, popMiddle
, and popBack
.请你设计一个队列,支持在前,中,后三个位置的 push
和 pop
操作。
请你完成 FrontMiddleBack
类:
FrontMiddleBack()
初始化队列。void pushFront(int val)
将 val
添加到队列的 最前面 。void pushMiddle(int val)
将 val
添加到队列的 正中间 。void pushBack(int val)
将 val
添加到队里的 最后面 。int popFront()
将 最前面 的元素从队列中删除并返回值,如果删除之前队列为空,那么返回 -1
。int popMiddle()
将 正中间 的元素从队列中删除并返回值,如果删除之前队列为空,那么返回 -1
。int popBack()
将 最后面 的元素从队列中删除并返回值,如果删除之前队列为空,那么返回 -1
。请注意当有 两个 中间位置的时候,选择靠前面的位置进行操作。比方说:
\n\n6
添加到 [1, 2, 3, 4, 5]
的中间位置,结果数组为 [1, 2, 6, 3, 4, 5]
。[1, 2, 3, 4, 5, 6]
的中间位置弹出元素,返回 3
,数组变为 [1, 2, 4, 5, 6]
。\n\n
示例 1:
\n\n\n输入:\n[\"FrontMiddleBackQueue\", \"pushFront\", \"pushBack\", \"pushMiddle\", \"pushMiddle\", \"popFront\", \"popMiddle\", \"popMiddle\", \"popBack\", \"popFront\"]\n[[], [1], [2], [3], [4], [], [], [], [], []]\n输出:\n[null, null, null, null, null, 1, 3, 4, 2, -1]\n\n解释:\nFrontMiddleBackQueue q = new FrontMiddleBackQueue();\nq.pushFront(1); // [1]\nq.pushBack(2); // [1, 2]\nq.pushMiddle(3); // [1, 3, 2]\nq.pushMiddle(4); // [1, 4, 3, 2]\nq.popFront(); // 返回 1 -> [4, 3, 2]\nq.popMiddle(); // 返回 3 -> [4, 2]\nq.popMiddle(); // 返回 4 -> [2]\nq.popBack(); // 返回 2 -> []\nq.popFront(); // 返回 -1 -> [] (队列为空)\n\n\n
\n\n
提示:
\n\n1 <= val <= 109
1000
次 pushFront
, pushMiddle
, pushBack
, popFront
, popMiddle
和 popBack
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
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