Problem

Implement a function to check if a given linked list is a palindrome.
Input: 1 -> 2 -> 3 -> 2 -> 1
Output: True

Solution

# Solution:
# The solution is to reverse the linked list and compare it to the original.
# This is optimal because it only requires one pass through the linked list.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, data):
        new_node = Node(data)
        if self.head is None:
            self.head = new_node
            return
        last_node = self.head
        while last_node.next:
            last_node = last_node.next
        last_node.next = new_node

    def print_list(self):
        cur_node = self.head
        while cur_node:
            print(cur_node.data)
            cur_node = cur_node.next

    def reverse(self):
        prev = None
        cur = self.head
        while cur:
            nxt = cur.next
            cur.next = prev
            prev = cur
            cur = nxt
        self.head = prev

    def is_palindrome(self):
        self.reverse()
        cur = self.head
        while cur:
            if cur.data != self.head.data:
                return False
            cur = cur.next
            self.head = self.head.next
        return True

llist = LinkedList()
llist.append("R")
llist.append("A")
llist.append("D")
llist.append("A")
llist.append("R")
print(llist.is_palindrome())