
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);
}