Find midpoint, reverse, weave
Alternate Relay Ends
01. The question
Build a singly linked chain and relink it in first, last, second, next-to-last order, without changing any node’s reading. Return the new reading sequence; empty and single-node chains stay unchanged.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input[2,5,8,11,14]
Output[2,14,5,11,8]
- Slow/fast pointers split the chain after 8.
- Reverse the second half to 14 → 11.
- Weave 2 → 14 → 5 → 11 → 8, saving both next pointers before each splice.
Try your code
Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.
Edit your test cases
Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.
Input contract & reference code
Example data: [2,5,8,11,14]. Parameter: "". Numeric parameters arrive as strings; convert them with int(parameter) when needed.
The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.
Python downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.
Printed output
Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.
03. The idea
Try sketching your own approach first. What would you track, and why would it work?
Reveal the idea
- Locate the first half’s tail with slow and fast.
- Detach and reverse the second half using pointers.
- Alternate one node from each half, preserving saved successors.
04. Watch it happen
Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.
Input & settings
Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.
Working state / what changes
Ready to begin
Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.
Python / why it changes
The line behind the step
class Node:
def __init__(self, value):
self.value = value
self.next = None
def chain(values):
dummy = Node(None)
tail = dummy
for value in values:
tail.next = Node(value)
tail = tail.next
return dummy.next
def values(head):
result = []
while head is not None:
result.append(head.value)
head = head.next
return result
def solve(data, parameter=''):
head = chain(data)
if head is None or head.next is None:
return values(head)
slow = fast = head
while fast.next is not None and fast.next.next is not None:
slow = slow.next
fast = fast.next.next
current = slow.next
slow.next = None
previous = None
while current is not None:
following = current.next
current.next = previous
previous = current
current = following
first, second = head, previous
while second is not None:
next_first, next_second = first.next, second.next
first.next = second
second.next = next_first
first, second = next_first, next_second
return values(head)How to use this walkthrough
Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.
05. Python solution
class Node:
def __init__(self, value):
self.value = value
self.next = None
def chain(values):
dummy = Node(None)
tail = dummy
for value in values:
tail.next = Node(value)
tail = tail.next
return dummy.next
def values(head):
result = []
while head is not None:
result.append(head.value)
head = head.next
return result
def solve(data, parameter=''):
head = chain(data)
if head is None or head.next is None:
return values(head)
slow = fast = head
while fast.next is not None and fast.next.next is not None:
slow = slow.next
fast = fast.next.next
current = slow.next
slow.next = None
previous = None
while current is not None:
following = current.next
current.next = previous
previous = current
current = following
first, second = head, previous
while second is not None:
next_first, next_second = first.next, second.next
first.next = second
second.next = next_first
first, second = next_first, next_second
return values(head)06. Cost & edge cases
- Time
- O(n)
- Space
- O(1) algorithm workspace; O(n) boundary construction and returned JSON
- The first half keeps the extra middle node for odd lengths.
- Detach the halves before reversal to avoid a loop.
- Duplicate values do not collapse node identity.