6-3 Deque

6-3 Deque

分数 5

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!\n"); break; case pop: X = Pop(D); if ( X==ERROR ) printf("Deque is Empty!\n"); break; case inject: scanf("%d", &X); if (!Inject(X, D)) printf("Memory is Full!\n"); break; case eject: X = Eject(D); if ( X==ERROR ) printf("Deque is Empty!\n"); 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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

/*这题我忘记处理POP返回的指针了，卡了我四个小时！！！*/

#define shuiyunsheng 66666666666666

Deque CreateDeque()
{
    Deque head = (Deque)malloc(sizeof(struct DequeRecord));
    PtrToNode d = (PtrToNode)malloc(sizeof(struct Node));
    d->Next = d;
    d->Last = d;
    head->Front = d;
    head->Rear = d;
    return head;
}
int Push(ElementType X, Deque D)
{
    PtrToNode d = (PtrToNode)malloc(sizeof(struct Node));
    d->Element = X;
    if (D->Front == D->Rear)//如果为空,尾部指针需要更新
    {
        D->Rear = d;
    }
    d->Next = D->Front->Next;
    d->Last = D->Front;
    D->Front->Next->Last = d;
    D->Front->Next = d;
    return 1;
}
ElementType Pop(Deque D)
{
    if (D->Front == D->Rear)
        return ERROR;
    int n = D->Front->Next->Element;
    PtrToNode k = D->Front->Next;
    if (k == D->Rear)
    {
        D->Rear = D->Front;
    }
    k->Next->Last = D->Front;//Last 处理!!!!
    D->Front->Next = k->Next;
    free(k);
    return n;
}
int Inject(ElementType X, Deque D)
{
    PtrToNode d = (PtrToNode)malloc(sizeof(struct Node));
    d->Element = X;
    d->Next = D->Front;
    d->Last = D->Rear;
    D->Rear->Next = d;
    D->Rear = d;
    D->Front->Last = d;
    return 1;
}
ElementType Eject(Deque D)
{
    if (D->Front == D->Rear)
        return ERROR;
    int n = D->Rear->Element;
    PtrToNode k = D->Rear->Last;
    k->Next = D->Rear->Next;
    D->Front->Last = k;
    free(D->Rear);
    D->Rear = k;
    return n;
}