Easy
Given the head of a singly linked list, return true if it is a palindrome.
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, 105]. 0 <= Node.val <= 9
Follow up: Could you do it in
O(n) time and O(1) space? # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def get_head_mid(self, head):
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
def reverse_mid(self, head):
prev = None
current = head
while current:
temp = current.next
current.next = prev
prev = current
current = temp
return prev
def isPalindrome(self, head: Optional[ListNode]) -> bool:
head_first = head
head_mid = self.get_head_mid(head)
new_rev_head = self.reverse_mid(head_mid)
head_last = new_rev_head
while head_last:
if head_first.val != head_last.val:
return False
head_first = head_first.next
head_last = head_last.next
return True