从尾到头颠倒双向链接列表

Question

I am currently practicing pointers on my school break and below I written the method to reverse a doubly linked list, but when I hand it into an online test, it fails.

Node* Reverse(Node *head)
{
    int count = 0;
    struct Node *ptr = head;

    // return head if NULL
    if (head == NULL) {
        return head;
    }
    // if the list is only the head, then the reverse is just the head... so nothing changes
    if((head->next == NULL && head->prev == NULL)){
        return head;
    }

    //Come here if previous if statements fail, traverse the list until I reach tail which will become the
    // new head
    while(ptr->next != NULL){
        ptr = ptr->next;
        count++;
    }
    head = ptr; // this is the new head    
    //starting from tail all the way to head swap the "prev" and "next" of each node 
    struct Node *temp = ptr->next;

    for(int i = 0; i<count; i++){
        ptr->next = ptr->prev;
        ptr->prev = temp;
        ptr=ptr->next;
        temp= ptr->next;
        //count--;
    }

    return head;
}

I realize that it is probably smarter to reverse the list while I traverse it from head to tail, but I thought that was boring, so I decided to reverse it starting from the tail to head instead. I suspect there is an obvious error in my while loop or for loop, but I am unable to diagnose the error.

Answer 1

I think the error is here:

while(ptr->next != NULL){
    ptr = ptr->next;
    count++;
}

Let's say your linked list has 2 elements in it. Then that while loop will only iterate once, and count will be 1. When you get down to the for loop, it will also only iterate once, which means you will correctly reassign the pointers for the new head, but not the second element (previously the head).

If you initialize count to 1 instead of 0, it should correctly reflect the number of elements in the linked list and the for loop should execute correctly.

Edit: You will also have to restructure your for loop slightly to avoid a segfault at the end of the list:

Node* temp;

for (int i = 0; i < count; i++)
{
    temp = ptr->next;
    ptr->next = ptr->prev;
    ptr->prev = temp;
    ptr = ptr->next;
}

Answer 2

replace

for(int i = 0; i<count; i++){//i<count --> i<=count : Because Not counting last element
    ptr->next = ptr->prev;
    ptr->prev = temp;
    ptr=ptr->next;
    temp= ptr->next;//<-- bad
    //count--;
}

with

for(int i = 0; i <= count; i++){
    ptr->next = ptr->prev;
    ptr->prev = temp;
    temp = ptr;
    ptr = ptr->next;
}

or

while(ptr){
    ptr->next = ptr->prev;
    ptr->prev = temp;
    temp = ptr;//next prev
    ptr = ptr->next;
}