Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given `1->2->3->4->5->NULL`m = 2 and n = 4,

return `1->4->3->2->5->NULL`.

Note:
Given mn satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

(Java Code on github at the bottom of the post. )

My thoughts and solutions:

Well, I did a similar question before, but that experience didn’t come to me immediately. So this is how I proceeded:

1. When I see reverse, I think about the Stack structure.
2. Reversing a linked list does not necessarily mean moving the whole node around. It may just require you to reverse the value in the node.
3. Moving the node somehow to reverse the list.

The question explicitly requires one pass and in-place, so idea No.1 is a no, but still valuable thought for other problems. I moved on to idea No. 2 and I got stuck for while. So maybe it is not the right direction for this question. OK, so move on to idea  No.3. I kept asking myself this question: have you seen it before? Then suddenly something came up: I solved a question of reversing a whole list in-place and one pass before. And this question is just a more general version.

Say we have a list:

N1 -> N2 -> N3 -> NULL

If we reverse the whole list, it becomes:

N3 -> N2 -> N1 -> NULL

OK. to get N1 -> NULL is easy, we just set N1.next = NULL. What about N2 -> N1 ? The same, we set N2.next = N1. Wait! How do we get to N2? Eh, N2 = N1.next. But did we just set N1.next = NULL? So we won’t be able get N2 now, unless we save the reference to N2 somewhere else before setting N1.next = NULL.

```Node current = N1;
Node previous = NULL;

Node nextNode = current.next;
current.next = previous;
previous = current; //previous points to N1;
current = nextNode; //current points to N2;
```

I hope you get the idea. This code snippet should be wrapped in a loop to process the whole list. But knowing this does not make things easier. The corner cases can be tricky and it took me a while to make sure the answer was bug-free. Do it manually, not in any IDE.

Things to check:

• when there is only one node in the list.
• when m = 1
• when m == n

Code on github: