1670. Design Front Middle Back Queue #

题目 #

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:

Pushing 6 into the middle of [1, 2, 3, 4, 5] results in [1, 2, 6, 3, 4, 5].
Popping the middle from [1, 2, 3, 4, 5, 6] returns 3 and results in [1, 2, 4, 5, 6].

Example 1:

Input:
["FrontMiddleBackQueue", "pushFront", "pushBack", "pushMiddle", "pushMiddle", "popFront", "popMiddle", "popMiddle", "popBack", "popFront"]
[[], [1], [2], [3], [4], [], [], [], [], []]
Output:
[null, null, null, null, null, 1, 3, 4, 2, -1]

Explanation:
FrontMiddleBackQueue q = new FrontMiddleBackQueue();
q.pushFront(1);   // [1]
q.pushBack(2);    // [1, 2]
q.pushMiddle(3);  // [1, 3, 2]
q.pushMiddle(4);  // [1, 4, 3, 2]
q.popFront();     // return 1 -> [4, 3, 2]
q.popMiddle();    // return 3 -> [4, 2]
q.popMiddle();    // return 4 -> [2]
q.popBack();      // return 2 -> []
q.popFront();     // return -1 -> [] (The queue is empty)

Constraints:

1 <= val <= 109
At most 1000 calls will be made to pushFront, pushMiddle, pushBack, popFront, popMiddle, and popBack.

题目大意 #

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：

将 6 添加到 [1, 2, 3, 4, 5] 的中间位置，结果数组为 [1, 2, 6, 3, 4, 5] 。
从 [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] 。

解题思路 #

简单题，利用 go 原生的双向队列 list 的实现，可以轻松实现这个“前中后队列”。
具体实现见代码，几组特殊测试用例见测试文件。

代码 #

package leetcode

import (
	"container/list"
)

type FrontMiddleBackQueue struct {
	list   *list.List
	middle *list.Element
}

func Constructor() FrontMiddleBackQueue {
	return FrontMiddleBackQueue{list: list.New()}
}

func (this *FrontMiddleBackQueue) PushFront(val int) {
	e := this.list.PushFront(val)
	if this.middle == nil {
		this.middle = e
	} else if this.list.Len()%2 == 0 && this.middle.Prev() != nil {
		this.middle = this.middle.Prev()
	}
}

func (this *FrontMiddleBackQueue) PushMiddle(val int) {
	if this.middle == nil {
		this.PushFront(val)
	} else {
		if this.list.Len()%2 != 0 {
			this.middle = this.list.InsertBefore(val, this.middle)
		} else {
			this.middle = this.list.InsertAfter(val, this.middle)
		}
	}
}

func (this *FrontMiddleBackQueue) PushBack(val int) {
	e := this.list.PushBack(val)
	if this.middle == nil {
		this.middle = e
	} else if this.list.Len()%2 != 0 && this.middle.Next() != nil {
		this.middle = this.middle.Next()
	}
}

func (this *FrontMiddleBackQueue) PopFront() int {
	if this.list.Len() == 0 {
		return -1
	}
	e := this.list.Front()
	if this.list.Len() == 1 {
		this.middle = nil
	} else if this.list.Len()%2 == 0 && this.middle.Next() != nil {
		this.middle = this.middle.Next()
	}
	return this.list.Remove(e).(int)
}

func (this *FrontMiddleBackQueue) PopMiddle() int {
	if this.middle == nil {
		return -1
	}
	e := this.middle
	if this.list.Len()%2 != 0 {
		this.middle = e.Prev()
	} else {
		this.middle = e.Next()
	}
	return this.list.Remove(e).(int)
}

func (this *FrontMiddleBackQueue) PopBack() int {
	if this.list.Len() == 0 {
		return -1
	}
	e := this.list.Back()
	if this.list.Len() == 1 {
		this.middle = nil
	} else if this.list.Len()%2 != 0 && this.middle.Prev() != nil {
		this.middle = this.middle.Prev()
	}
	return this.list.Remove(e).(int)
}

/**
 * Your FrontMiddleBackQueue object will be instantiated and called as such:
 * obj := Constructor();
 * obj.PushFront(val);
 * obj.PushMiddle(val);
 * obj.PushBack(val);
 * param_4 := obj.PopFront();
 * param_5 := obj.PopMiddle();
 * param_6 := obj.PopBack();
 */

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