LeetCode：设计循环队列

题目链接：622. 设计循环队列 - 力扣（Leetcode）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);
*/

                                                              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

 

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