1670. Design Front Middle Back Queue

# 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 #

• 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 。

• 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();
*/
``````

Apr 8, 2023