交换链表中的节点

Question

I am trying to swap two adjacent nodes in a linked list, and I think I understand the idea of how to do it using a temporary node.

Here is my struct swap function

struct part {
   char* name;
   float price;
   int quantity;
   struct part *next;
};
typedef struct part partType;

partType *swap_node(partType **item) {

  partType *temp;
  temp = *item;
  *item = (*item)->next;
  temp->next = (*item)->next;
  (*item)->next = temp;
  return *item;
}

I cant think of how to make the previous node in the list point to the new swapped node. Do i need another temp variable? Also, how do I account for the case that the two nodes to be swapped are the first two in the list.

Answer 1

From the code, it looks like you want to swap item and item->next.

If you don't have a doubly-linked list, then you need to set linkPtr to head, and then iterate until linkPtr->next == *item. From there, you can start switching between linkPtr, linkPtr->next and linkPtr->next->next.

You also need a separate condition comparing linkPtr to head, and if so, then you need to set head to the new head.

Answer 2

Ignore the answers about doubly-linked lists. To answer your question, you need to think about how you call your function.

Right now, you have a function that takes a pointer to a pointer. It currently points to a node (node A), which in turn points to another node (node B). Imagine this scenario:

partType a, b, c, d;
a->next = &b;
b->next = &c;
c->next = &d;
d->next = NULL;

Now, you want to swap the order of B and C to have A->C->B->D using your function. Well, you'd do:

swap_node(&a->next);

A was pointing to B; now it's pointing to C. As you can see, the "previous" node is already pointing to C, as you expected. In other words, you've already accomplished your goal. Cheers!

Notes: What exactly is happening in your swap function? Let's break it down. First, the parameter you give it is a pointer to a pointer . Those are a bitch to think about because of the wording -- don't let the wording fool you. Just like "rate of change of the rate of change" is a bitch to think about but "acceleration" is much easier. You want to parse it by remembering that the parameter is, first and foremost, a pointer to some data, and your function is going to modify the data that it points to.

So your function gets a pointer to this 'p', which is pointing to a spot in the linked list which (you assume, see PS) points to two nodes (call them X and Y). Diagram:

[p] --> X[next] --> Y[next] --> Z[next]

Your algorithm does:

1. Make [p] point to Y: *item = (*item)->next
2. Make X[next] point to Z: temp->next = (*item)->next
3. Make Y[next] point to X: (*item)->next = temp

So, if you now consider my A, B, C, D example, the linked list was:

A[next] --> B[next] --> C[next] --> D[next] --> NULL

you can see more clearly what pointer I'm passing. It's the location in memory (read: pointer) where A[next] is stored, which your function needs to do the swapping.

Incidentally, another way to code this would be to do:

a->next = swap_node(&a->next);

but don't do that. It's redundant.

PS Have you thought about what happens when you ask to swap the last node in the series? Right now, things explode :P

Answer 3

With data this small, you might as well just swap everything but the next pointers:

partType tmp = *item;
memcpy(item, item->next, offsetof(item, next));
memcpy(item->next, &tmp, offsetof(item, next));

If your data gets too large to do this, you'll need a pointer to the node before the two you want. The nice part is that your fixing of prev 's next pointer acts as a temp variable, letting you not need one.

prev->next = item->next;
item->next = item->next->next;
prev->next->next = item;

Answer 4

Simple way to swap to node... I am not writing actual code. I am just giving you a hint to swap nodes.

[1]->[2]->[3]->[4]

Suppose this is your linked list and you want to swap [2] and [3] .

1. use loop to reach till [2] . so your temp is at [2] .
2. Now temp1 = temp->next; Hence temp1 is at [3] .

3. temp->next = temp1->next;

   temp1->next = temp;

so now temp->next = [4] and temp1->next = [2]

Answer 5

Four easy options:

First option: Keep track of the pointer to prev (probably the direction the assignment is expected to take).

Second option: Implement a doubly linked list so that you can always find the previous node.

Third option: Implement a singly linked list (as your assignment probably calls for), but give each node two pointers: One pointing to 'next', and one pointing to the data payload (which could be just about anything; a struct, an array, or a plain old data type). Then you just swap where the payload pointers point without worrying about 'next' and 'prev'.

A fourth option: Swap the data itself (possibly the direction the assignment is expected to take).

There are use cases for each of the above.