easyDSA

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

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

**Intuition.** You can't index backwards in a singly linked list, so reverse the *second half* and compare it node-by-node with the first half. Find the middle with slow/fast pointers, reverse from there, then walk the two halves inward. O(n) time, O(1) space. 1→2→2→1: reverse 2nd half, compare with 1st half 1 2

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Explain the reverse-half approach.

linked-listtwo-pointers

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Reverse second half

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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

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