Reorder List

To solve this coding challenge, we need to reorder the given singly linked-list as per the specified pattern while ensuring that only the nodes themselves can be changed, not the values within them. The reordering involves:

• Splitting the list into two halves.
• Reversing the second half.
• Merging the two halves in the prescribed order.

Here’s a step-by-step method to achieve the solution:

# Explanation

1. Find the middle of the linked list: Use the slow and fast pointer technique. Slow pointer moves one step at a time while the fast pointer moves two steps. When the fast pointer reaches the end, the slow pointer will be at the middle.
2. Reverse the second half of the linked list: Starting from the middle to the end, reverse the linked list to facilitate easy merging.
3. Merge the two halves: Merge the first half and the reversed second half so that the nodes alternate from each half as per the requirement.

Detailed Steps in Pseudocode

4. Initialization & Edge Case Handling:
• If the list is empty or has only one node, no reordering is needed.
5. Find the Middle:
• Initialize two pointers,
  slow
  and
  fast
  , both pointing to the head of the list.
• Move
  slow
  one step and
  fast
  two steps at a time until
  fast
  reaches the end.
  Slow
  will be at the middle of the list.
6. Reverse the Second Half:
• Initialize two pointers,
  prev
  (set to
  None
  ) and
  curr
  (set to
  slow
  ).
• Traverse from the middle to the end of the list, reversing the pointers (i.e.,
  curr.next = prev
  ) at each step until the end of the list is reached.
7. Merge the Two Halves:
• Initialize two pointers,
  first
  (set to
  head
  ) and
  second
  (set to
  prev
  ).
• Alternate between advancing the
  first
  and
  second
  pointers and linking them until all nodes are connected in the required order.

# Pseudocode

                                            
# Function to reorder the list
function reorderList(head)

# Edge case: check if the list is empty or has a single node
if head is NULL or head.next is NULL
return

# Step 1: Find the middle of the linked list
slow = head
fast = head
while fast is not NULL and fast.next is not NULL
slow = slow.next
fast = fast.next.next

# Now, slow is at the middle node

# Step 2: Reverse the second half of the list
prev = NULL
curr = slow
while curr is not NULL
next_node = curr.next
curr.next = prev
prev = curr
curr = next_node

# Now, prev is the head of the reversed second half

# Step 3: Merge the two halves
first = head
second = prev
while second.next is not NULL
temp1 = first.next
first.next = second
first = temp1

temp2 = second.next
second.next = first
second = temp2

# List is now reordered according to the problem's requirements

end function

                                        

A visual rundown of these steps with an example list
[1, 2, 3, 4]
would be:
8. The
   slow
   pointer will end up at node
   3
   and
   fast
   at the end.

                                            
1 -> 2 -> 3 -> 4
^         ^
slow      fast

                                        

9. Reverse the list from the
   slow
   node:

                                            
1 -> 2 -> 3 <- 4
^    ^
slow prev

                                        

10. Merge the two halves:

                                            
1 -> 4 -> 2 -> 3

                                        

This solution follows a time complexity of O(n) and a space complexity of O(1), making it efficient given the constraints.