Explain the reverse-half approach.
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easyDSA
How do you check if a linked list is a palindrome?
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01
Understand the problem
linked-listtwo-pointers
02
Attempt it yourself
Sketch your approach before reading the solution — that's what interviews test.
Nudge consolestandby
Stuck? Beam a request up — the console returns a conceptual nudge that guides your logic without spoiling the implementation.
03
Study the solution
Find the middle with slow/fast pointers, reverse the second half, then compare it node-by-node with the first half. Optionally restore the list afterward. O(n) time and O(1) space, beating the O(n)-space copy-to-array approach.
Solution ready — 2 min read
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04
Read the code
Reverse second half
Run Playgroundclass ListNode:
def __init__(self, val, nxt=None):
self.val, self.next = val, nxt
def is_palindrome(head):
slow = fast = head # find middle
while fast and fast.next:
slow = slow.next
fast = fast.next.next
prev = None # reverse second half
while slow:
slow.next, prev, slow = prev, slow, slow.next
left, right = head, prev # compare
while right:
if left.val != right.val:
return False
left, right = left.next, right.next
return True
# --- demo ---
def build(vals):
head = None
for v in reversed(vals): head = ListNode(v, head)
return head
print(is_palindrome(build([1, 2, 2, 1]))) # True
print(is_palindrome(build([1, 2, 3]))) # False05
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