[数据结构]设计循环队列

醉城夜风~2025-03-06 8:15
思路展示：
代码实现：
复制代码
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);
}