思路展示：

代码实现：

typedef struct {int *a;int front;int rear;int k;
} MyCircularQueue;bool myCircularQueueIsEmpty(MyCircularQueue* obj);bool myCircularQueueIsFull(MyCircularQueue* obj);
MyCircularQueue* myCircularQueueCreate(int k) {MyCircularQueue* obj=(MyCircularQueue*)malloc(sizeof(MyCircularQueue));obj->front=obj->rear=0;obj->a=(int *)malloc(sizeof(int)*(k+1));obj->k=k;return obj;
}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)//取第一个元素
{if(myCircularQueueIsEmpty(obj)){return -1;}else{return obj->a[obj->front];}
}int myCircularQueueRear(MyCircularQueue* obj) {//return obj->a[obj->rear-1];  不能这样访问，如果这样访问那么当rear移动到开头那么rear-1就是-1就是越界访问了if( myCircularQueueIsEmpty(obj)){return -1;}else{return obj->a[(obj->rear-1+obj->k+1)%(obj->k+1)];//如果rear-1>0那么rear-1+k再模k是不变的}
}bool myCircularQueueIsEmpty(MyCircularQueue* obj) {return obj->front==obj->rear;
}bool myCircularQueueIsFull(MyCircularQueue* obj) {return (obj->rear+1)%(obj->k+1)==obj->front;
}void myCircularQueueFree(MyCircularQueue* obj) {free(obj->a);free(obj);
}