Go to “Reorder List (#143)” on Leetcode ↗

This problem requires making a couple of observations and small conceptual jumps:

  • The reordering involves interleaving two “streams” of nodes, one going in the direction of the original linked list and one going in the opposite direction.
  • This seems like it would be fairly easy if we had a second copy of the linked list, but in reverse order. Can we make one? Well, we can use a stack to reverse a linked list — we can traverse through the linked list, adding each element to the stack, and we can pop items off the stack; they’ll come off in reverse order.
  • The problem asks us to modify the linked list in-place, but modifying in-place can sometimes be messy while you’re in the middle of it. Sometimes it’s easier to imagine building a new data structure from scratch, and figuring out how to make that work in-place. So how would we build a new, reordered linked list? We could alternate between the two lists, getting the head of the linked list and popping the top item of the stack, and adding them to our new linked list one at a time.

In this case, it was important to notice the specific pattern in the reordered list — the pattern isn’t arbitrary, and its design informs the solution. You also need to know (or figure out) that a linked list can only be traversed in one direction, and a items popped off a stack come off in the opposite order that they were added in; combining these facts means that a stack can be used to reverse a linked list. Finally, you’ll have to combine these pieces into the actual solution.

There are two remaining questions:

How do we adapt the solution to modify the original list in-place? Well, we want to keep the first head in place, but we want to insert a node between that head and the head.next node. So, we can stash head.next in a temporary variable, pop the node we want to insert-in-between from the stack, set head.next to that node, and set its next to the temporary variable. Then we repeat, with head now referring to the node in the temporary variable.

Finally, how do we know when we’re done? If we manually run that solution over the examples, we see that we have to run those steps N/2 steps, where N is the number of items in the linked list. What happens if N is odd? In Example 2, N is 5, and it turns out that running these steps 2 times is also enough. Thus, N/2 (rounded down, which is what integer division does by default) is the number of times we need to run these steps.

class ListNode
  attr_accessor :val, :next

  def initialize(val)
    @val = val
    @next = nil

def reorder_list(head)
  # Create stack from given linked list
  stack = []
  node = head
  node_count = 0
  while node
    # Also count the number of nodes
    # since we're going through all of them anyway
    node_count += 1
    node = node.next

  # Nothing to do if the given list is empty
  return nil if node_count == 0

  # Starting from `head`, replace the `next` bit of the list
  # by building a new one
  node_from_beginning = head
  (node_count / 2).times do
    subsequent_node = node_from_beginning.next
    node_from_end = stack.pop
    node_from_end.next = subsequent_node
    node_from_beginning.next = node_from_end
    node_from_beginning = subsequent_node
  node_from_beginning.next = nil

This solution iterates through all the nodes of the linked list once to build the stack — this is O(N). We’ll assume pushing and poping to a stack is constant time, so we can ignore this factor. Then, we run a bunch of constant-time updates within a loop which repeats N/2 times — this is O(N/2). Thus, the total runtime is O(N) + O(N/2), which simplifies to O(N).