LeetCode：设计循环队列

梦里梦外; 2023-10-10 18:58 1阅读 0赞

题目链接：[622. 设计循环队列 - 力扣（Leetcode）![icon-default.png_t_N4HB][]https://leetcode.cn/problems/design-circular-queue/][622. _ - _Leetcode_icon-default.png_t_N4HB_https_leetcode.cn_problems_design-circular-queue]

还是老套路二话不说，先上代码

typedef struct {
        int front;
        int rear;
        int k;
        int* a;
    } MyCircularQueue;
    
    MyCircularQueue* myCircularQueueCreate(int k) {
        MyCircularQueue* obj = (MyCircularQueue*)malloc(sizeof(MyCircularQueue));
        obj->a = (int*)malloc(sizeof(int)*(k+1));
        if(obj==NULL || obj->a == NULL)
        {
            perror("malloc fail");
            return NULL;
        }
        obj->k=k;
        obj->front = obj->rear =0;
        return obj;
    }
    bool myCircularQueueIsEmpty(MyCircularQueue* obj) {
        return obj->front == obj->rear;
    }
    bool myCircularQueueIsFull(MyCircularQueue* obj) {
        return (obj->rear+1) % (obj->k+1) == obj->front;
    }
    bool myCircularQueueEnQueue(MyCircularQueue* obj, int value) {
        if(myCircularQueueIsFull(obj))
            return false;
        obj->a[obj->rear] = value;
        obj->rear++;
        obj->rear %= (obj->k+1);
        return true;
    }
    bool myCircularQueueDeQueue(MyCircularQueue* obj) {
        if(myCircularQueueIsEmpty(obj))
            return false;
        obj->front++;
        obj->front %= (obj->k+1);
        return true;
    }
    int myCircularQueueFront(MyCircularQueue* obj) {
        if(myCircularQueueIsEmpty(obj))
            return -1;
        return obj->a[obj->front];
    }
    
    int myCircularQueueRear(MyCircularQueue* obj) {
        if(myCircularQueueIsEmpty(obj))
            return -1;
        return obj->a[(obj->rear+obj->k) % (obj->k + 1)];
    }
    
    
    
    void myCircularQueueFree(MyCircularQueue* obj) {
        free(obj->a);
        free(obj);
    }
    
    /**
     * Your MyCircularQueue struct will be instantiated and called as such:
     * MyCircularQueue* obj = myCircularQueueCreate(k);
     * bool param_1 = myCircularQueueEnQueue(obj, value);
     
     * bool param_2 = myCircularQueueDeQueue(obj);
     
     * int param_3 = myCircularQueueFront(obj);
     
     * int param_4 = myCircularQueueRear(obj);
     
     * bool param_5 = myCircularQueueIsEmpty(obj);
     
     * bool param_6 = myCircularQueueIsFull(obj);
     
     * myCircularQueueFree(obj);
    */

![cea5477682b346e1a231c28a8209280c.png][]

**过啦！！！！！！**

--------------------

解题思路：

>     通过一个定长数组实现循环队列
>     入队：首先要判断队列是否已满，再进行入队的操作，入队操作需要考虑索引循环的问题，当索引越界，需要让它变成最小值
>     出队：首先要判断队列是否为空，再进行出队操作，出队也需要考虑索引循环的问题
>     判空： 队头 == 队尾
>     判满： 队尾 + 1 == 队头

![b68c152bf9074513aa27bbf4542af021.png][]

**实现方式：**

**假设循环队列储存K个数据，则要开辟K+1的空间**

![277245db4d67419084e3d3b7491ef839.png][]

![44acaf8611d14f3d9cb5cb21fb5d37fd.png][]

[icon-default.png_t_N4HB]: https://csdnimg.cn/release/blog_editor_html/release2.2.8/ckeditor/plugins/CsdnLink/icons/icon-default.png?t=N4HB
[622. _ - _Leetcode_icon-default.png_t_N4HB_https_leetcode.cn_problems_design-circular-queue]: https://leetcode.cn/problems/design-circular-queue/
[cea5477682b346e1a231c28a8209280c.png]: https://img-blog.csdnimg.cn/cea5477682b346e1a231c28a8209280c.png
[b68c152bf9074513aa27bbf4542af021.png]: https://img-blog.csdnimg.cn/b68c152bf9074513aa27bbf4542af021.png
[277245db4d67419084e3d3b7491ef839.png]: https://img-blog.csdnimg.cn/277245db4d67419084e3d3b7491ef839.png
[44acaf8611d14f3d9cb5cb21fb5d37fd.png]: https://img-blog.csdnimg.cn/44acaf8611d14f3d9cb5cb21fb5d37fd.png