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https://github.com/halfrost/LeetCode-Go.git
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Add solution 0622
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package leetcode
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type MyCircularQueue struct {
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cap int
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size int
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queue []int
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left int
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right int
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}
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func Constructor(k int) MyCircularQueue {
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return MyCircularQueue{cap: k, size: 0, left: 0, right: 0, queue: make([]int, k)}
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}
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func (this *MyCircularQueue) EnQueue(value int) bool {
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if this.size == this.cap {
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return false
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}
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this.size++
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this.queue[this.right] = value
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this.right++
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this.right %= this.cap
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return true
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}
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func (this *MyCircularQueue) DeQueue() bool {
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if this.size == 0 {
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return false
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}
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this.size--
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this.left++
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this.left %= this.cap
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return true
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}
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func (this *MyCircularQueue) Front() int {
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if this.size == 0 {
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return -1
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}
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return this.queue[this.left]
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}
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func (this *MyCircularQueue) Rear() int {
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if this.size == 0 {
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return -1
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}
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if this.right == 0 {
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return this.queue[this.cap-1]
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}
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return this.queue[this.right-1]
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}
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func (this *MyCircularQueue) IsEmpty() bool {
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return this.size == 0
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}
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func (this *MyCircularQueue) IsFull() bool {
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return this.size == this.cap
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}
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/**
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* Your MyCircularQueue object will be instantiated and called as such:
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* obj := Constructor(k);
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* param_1 := obj.EnQueue(value);
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* param_2 := obj.DeQueue();
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* param_3 := obj.Front();
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* param_4 := obj.Rear();
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* param_5 := obj.IsEmpty();
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* param_6 := obj.IsFull();
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*/
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@ -0,0 +1,23 @@
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package leetcode
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import (
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"fmt"
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"testing"
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)
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func Test_Problem622(t *testing.T) {
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obj := Constructor(3)
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fmt.Printf("obj = %v\n", obj)
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param1 := obj.EnQueue(1)
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fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
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param2 := obj.DeQueue()
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fmt.Printf("param_1 = %v obj = %v\n", param2, obj)
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param3 := obj.Front()
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fmt.Printf("param_1 = %v obj = %v\n", param3, obj)
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param4 := obj.Rear()
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fmt.Printf("param_1 = %v obj = %v\n", param4, obj)
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param5 := obj.IsEmpty()
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fmt.Printf("param_1 = %v obj = %v\n", param5, obj)
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param6 := obj.IsFull()
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fmt.Printf("param_1 = %v obj = %v\n", param6, obj)
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}
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147
leetcode/0622.Design-Circular-Queue/README.md
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147
leetcode/0622.Design-Circular-Queue/README.md
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# [622. Design Circular Queue](https://leetcode.com/problems/design-circular-queue/)
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## 题目
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Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle and the last position is connected back to the first position to make a circle. It is also called "Ring Buffer".
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One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values.
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Implementation the `MyCircularQueue` class:
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- `MyCircularQueue(k)` Initializes the object with the size of the queue to be `k`.
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- `int Front()` Gets the front item from the queue. If the queue is empty, return `1`.
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- `int Rear()` Gets the last item from the queue. If the queue is empty, return `1`.
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- `boolean enQueue(int value)` Inserts an element into the circular queue. Return `true` if the operation is successful.
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- `boolean deQueue()` Deletes an element from the circular queue. Return `true` if the operation is successful.
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- `boolean isEmpty()` Checks whether the circular queue is empty or not.
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- `boolean isFull()` Checks whether the circular queue is full or not.
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**Example 1:**
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```
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Input
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["MyCircularQueue", "enQueue", "enQueue", "enQueue", "enQueue", "Rear", "isFull", "deQueue", "enQueue", "Rear"]
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[[3], [1], [2], [3], [4], [], [], [], [4], []]
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Output
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[null, true, true, true, false, 3, true, true, true, 4]
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Explanation
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MyCircularQueue myCircularQueue = new MyCircularQueue(3);
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myCircularQueue.enQueue(1); // return True
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myCircularQueue.enQueue(2); // return True
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myCircularQueue.enQueue(3); // return True
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myCircularQueue.enQueue(4); // return False
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myCircularQueue.Rear(); // return 3
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myCircularQueue.isFull(); // return True
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myCircularQueue.deQueue(); // return True
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myCircularQueue.enQueue(4); // return True
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myCircularQueue.Rear(); // return 4
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```
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**Constraints:**
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- `1 <= k <= 1000`
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- `0 <= value <= 1000`
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- At most `3000` calls will be made to `enQueue`, `deQueue`, `Front`, `Rear`, `isEmpty`, and `isFull`.
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**Follow up:**
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Could you solve the problem without using the built-in queue?
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## 题目大意
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设计你的循环队列实现。 循环队列是一种线性数据结构,其操作表现基于 FIFO(先进先出)原则并且队尾被连接在队首之后以形成一个循环。它也被称为“环形缓冲器”。
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循环队列的一个好处是我们可以利用这个队列之前用过的空间。在一个普通队列里,一旦一个队列满了,我们就不能插入下一个元素,即使在队列前面仍有空间。但是使用循环队列,我们能使用这些空间去存储新的值。
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你的实现应该支持如下操作:
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- MyCircularQueue(k): 构造器,设置队列长度为 k 。
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- Front: 从队首获取元素。如果队列为空,返回 -1 。
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- Rear: 获取队尾元素。如果队列为空,返回 -1 。
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- enQueue(value): 向循环队列插入一个元素。如果成功插入则返回真。
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- deQueue(): 从循环队列中删除一个元素。如果成功删除则返回真。
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- isEmpty(): 检查循环队列是否为空。
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- isFull(): 检查循环队列是否已满。
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## 解题思路
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- 简单题。设计一个环形队列,底层用数组实现。额外维护 4 个变量,队列的总 cap,队列当前的 size,前一元素下标 left,后一个元素下标 right。每添加一个元素便维护 left,right,size,下标需要对 cap 取余,因为超过 cap 大小之后,需要循环存储。代码实现没有难度,具体sh见下面代码。
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## 代码
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```go
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package leetcode
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type MyCircularQueue struct {
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cap int
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size int
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queue []int
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left int
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right int
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}
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func Constructor(k int) MyCircularQueue {
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return MyCircularQueue{cap: k, size: 0, left: 0, right: 0, queue: make([]int, k)}
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}
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func (this *MyCircularQueue) EnQueue(value int) bool {
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if this.size == this.cap {
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return false
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}
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this.size++
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this.queue[this.right] = value
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this.right++
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this.right %= this.cap
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return true
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}
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func (this *MyCircularQueue) DeQueue() bool {
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if this.size == 0 {
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return false
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}
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this.size--
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this.left++
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this.left %= this.cap
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return true
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}
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func (this *MyCircularQueue) Front() int {
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if this.size == 0 {
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return -1
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}
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return this.queue[this.left]
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}
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func (this *MyCircularQueue) Rear() int {
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if this.size == 0 {
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return -1
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}
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if this.right == 0 {
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return this.queue[this.cap-1]
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}
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return this.queue[this.right-1]
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}
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func (this *MyCircularQueue) IsEmpty() bool {
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return this.size == 0
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}
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func (this *MyCircularQueue) IsFull() bool {
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return this.size == this.cap
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}
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/**
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* Your MyCircularQueue object will be instantiated and called as such:
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* obj := Constructor(k);
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* param_1 := obj.EnQueue(value);
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* param_2 := obj.DeQueue();
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* param_3 := obj.Front();
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* param_4 := obj.Rear();
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* param_5 := obj.IsEmpty();
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* param_6 := obj.IsFull();
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*/
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```
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