A "deque" is a data structure consisting of a list of items, on which the following operations are possible:
- Push(X,D): Insert item X on the front end of deque D.
- Pop(D): Remove the front item from deque D and return it.
- Inject(X,D): Insert item X on the rear end of deque D.
- Eject(D): Remove the rear item from deque D and return it. Write routines to support the deque that take O(1) time per operation.
Format of functions:
Deque CreateDeque();
int Push( ElementType X, Deque D );
ElementType Pop( Deque D );
int Inject( ElementType X, Deque D );
ElementType Eject( Deque D );
where Deque
is defined as the following:
typedef struct Node *PtrToNode;
struct Node {
ElementType Element;
PtrToNode Next, Last;
};
typedef struct DequeRecord *Deque;
struct DequeRecord {
PtrToNode Front, Rear;
};
Here the deque is implemented by a doubly linked list with a header. Front
and Rear
point to the two ends of the deque respectively. Front
always points to the header. The deque is empty when Front
and Rear
both point to the same dummy header. Note: Push
and Inject
are supposed to return 1 if the operations can be done successfully, or 0 if fail. If the deque is empty, Pop
and Eject
must return ERROR
which is defined by the judge program.
Sample program of judge:
#include <stdio.h>
#include <stdlib.h>
#define ElementType int
#define ERROR 1e5
typedef enum { push, pop, inject, eject, end } Operation;
typedef struct Node *PtrToNode;
struct Node {
ElementType Element;
PtrToNode Next, Last;
};
typedef struct DequeRecord *Deque;
struct DequeRecord {
PtrToNode Front, Rear;
};
Deque CreateDeque();
int Push( ElementType X, Deque D );
ElementType Pop( Deque D );
int Inject( ElementType X, Deque D );
ElementType Eject( Deque D );
Operation GetOp(); /* details omitted */
void PrintDeque( Deque D ); /* details omitted */
int main()
{
ElementType X;
Deque D;
int done = 0;
D = CreateDeque();
while (!done) {
switch(GetOp()) {
case push:
scanf("%d", &X);
if (!Push(X, D)) printf("Memory is Full!
");
break;
case pop:
X = Pop(D);
if ( X==ERROR ) printf("Deque is Empty!
");
break;
case inject:
scanf("%d", &X);
if (!Inject(X, D)) printf("Memory is Full!
");
break;
case eject:
X = Eject(D);
if ( X==ERROR ) printf("Deque is Empty!
");
break;
case end:
PrintDeque(D);
done = 1;
break;
}
}
return 0;
}
/* Your function will be put here */
Sample Input:
Pop
Inject 1
Pop
Eject
Push 1
Push 2
Eject
Inject 3
End
Sample Output:
Deque is Empty!
Deque is Empty!
Inside Deque: 2 3
题意:这道题要求设计一个链式双端队列 deque,并且要求有头指针。
初始时,链表为空,头指针 Front 和尾指针 Rear 指向同一片空的空间,
Push :将项 X 插入双端队列 D 的前端。下图为 Push(2) Push(1) 操作结果
Pop:从双端队列 D 中删除前端一项并返回。下图为 Pop() 操作结果
Inject:将项目 X 插入双端队列 D 的后端。下图为 Inject(3) 操作结果
Eject:从双端队列 D 中删除后端一项并返回。
下图为Eject
() 操作结果
若 Push
和 Inject
操作成功,返回1,否则返回0;若双端队列为空,Pop
和 Eject
返回 ERROR,否则返回弹出结点的 Element。
代码:
Deque CreateDeque() {
PtrToNode temnode = (PtrToNode)malloc(sizeof(struct Node));
temnode->Next = temnode->Last = NULL;
Deque D = (Deque)malloc(sizeof(struct DequeRecord));
D->Front = D->Rear = temnode;
return D;
}
int Push(ElementType X, Deque D) {
PtrToNode temnode= (PtrToNode)malloc(sizeof(struct Node));
temnode->Element = X;
temnode->Next = temnode->Last = NULL;
if (D->Front == D->Rear)
D->Rear = temnode;
else {
temnode->Next = D->Front->Next;
D->Front->Next->Last = temnode;
}
D->Front->Next = temnode;
temnode->Last = D->Front;
return 1;
}
ElementType Pop(Deque D) {
if (D->Front == D->Rear)
return ERROR;
PtrToNode delnode = (PtrToNode)malloc(sizeof(struct Node));
delnode = D->Front->Next;
if (delnode->Next == NULL)
D->Rear = D->Front;
else{
delnode->Next->Last = D->Front;
D->Front->Next = delnode->Next;
}
return delnode->Element;
free(delnode);
}
int Inject(ElementType X, Deque D) {
PtrToNode temnode = (PtrToNode)malloc(sizeof(struct Node));
temnode->Element = X;
temnode->Next = temnode->Last = NULL;
if (D->Front == D->Rear) {
D->Front->Next = temnode;
temnode->Last = D->Front;
}
else {
D->Rear->Next = temnode;
temnode->Last = D->Rear;
}
D->Rear = temnode;
return 1;
}
ElementType Eject(Deque D) {
if (D->Front == D->Rear)
return ERROR;
PtrToNode delnode = (PtrToNode)malloc(sizeof(struct Node));
delnode = D->Rear;
if (delnode->Last == D->Front)
D->Rear = D->Front;
else
D->Rear = delnode->Last;
return delnode->Element;
free(delnode);
}