PalindromeLinkedList
Problem
https://leetcode.com/problems/palindrome-linked-list/
Given the head of a singly linked list, return true if it is a
palindrome or false otherwise.
Example 1:
Input: head = [1,2,2,1]
Output: true
Example 2:
Input: head = [1,2]
Output: false
Constraints:
The number of nodes in the list is in the range
[1, 10:sup:`5`].0 <= Node.val <= 9
Follow up: Could you do it in O(n) time and O(1) space?
Solution
Save the contents of the linked list in a regular list list. Then, use two
pointers``i`` and j to check if list is a palindrome, moving them
inwards if list[i] == list[j].
Pattern
Linked List, Two Pointers, Stack, Recursion
Code
from __future__ import annotations
from typing import List
class ListNode:
"""Node in a linked list.
"""
def __init__(self, val: int, next: ListNode | None = None):
self.val = val
self.next = next
@classmethod
def from_list(cls, list: List[int]) -> ListNode | None:
head: ListNode | None = None
for el in list:
if head is None:
head = ListNode(el)
node = head
else:
node.next = ListNode(el)
node = node.next
return head
def isPalindrome(head: ListNode | None) -> bool:
"""Checks if the linked list is a palindrome.
"""
list = []
while head is not None:
list.append(head.val)
head = head.next
i = 0
j = len(list) - 1
while i < j:
if list[i] != list[j]:
return False
i += 1
j -= 1
return True
Test
>>> from PalindromeLinkedList import ListNode, isPalindrome
>>> head = ListNode.from_list([1, 2, 2, 1])
>>> isPalindrome(head)
True
>>> head = ListNode.from_list([1, 2])
>>> isPalindrome(head)
False