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How do you check if a linked list is a palindrome?

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01

Understand the problem

Explain the reverse-half approach.

linked-listtwo-pointers
02

Attempt it yourself

Sketch your approach before reading the solution — that's what interviews test.

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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 Playground
class 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])))      # False
05

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