6-4 Reverse Linked List(20点)

6-4 Reverse Linked List(20点)
Write a nonrecursive procedure to reverse a singly linked list in O(N) time using constant extra space.
Format of functions:

List Reverse( List L );

where  List  is defined as the following:

typedef struct Node *PtrToNode;
typedef PtrToNode List;
typedef PtrToNode Position;
struct Node {
    ElementType Element;
    Position Next;
};

The function  Reverse  is supposed to return the reverse linked list of  L , with a dummy header.
Sample program of judge:

#include 
#include 

typedef int ElementType;
typedef struct Node *PtrToNode;
typedef PtrToNode List;
typedef PtrToNode Position;
struct Node {
    ElementType Element;
    Position Next;
};

List Read(); /* details omitted */
void Print( List L ); /* details omitted */
List Reverse( List L );

int main()
{
    List L1, L2;
    L1 = Read();
    L2 = Reverse(L1);
    Print(L1);
    Print(L2);
    return 0;
}

/* Your function will be put here */

Sample Input:

5
1 3 4 5 2

Sample Output:

2 5 4 3 1
2 5 4 3 1

List Reverse(List L){
if (L->Next == NULL || L->Next->Next==NULL)
return L;
else {
PtrToNode p = L->Next;
PtrToNode q = p->Next;
if (q->Next == NULL) {
p->Next = NULL;
q->Next = p;
}
else{
PtrToNode tmp = q->Next;
p->Next = NULL;
while (tmp) {
q->Next = p;
p = q;
q = tmp;
tmp = tmp->Next;
}
q->Next = p;
}
L->Next = q;
}
return L;
}