Description
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, 105]
. 0 <= Node.val <= 9
Follow up: Could you do it in
O(n)
time and O(1)
space?
Solution
Python3
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def isPalindrome(self, head: Optional[ListNode]) -> bool:
slow, fast = head, head
# 1) move slow pointer to the middle of the linked list
while fast and fast.next:
slow = slow.next
fast = fast.next.next
def reverseLL(curr):
prev = None
while curr:
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
return prev
# 2) reverse the second half of the linked list
rhead = reverseLL(slow)
# 3) compare the first half and the reversed second half
while rhead:
if head.val != rhead.val:
return False
head = head.next
rhead = rhead.next
return True
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
bool isPalindrome(ListNode* head) {
if (head == nullptr || head->next == nullptr) return true;
ListNode *slow = head, *fast = head;
while (fast->next && fast->next->next){
slow = slow->next;
fast = fast->next->next;
}
slow->next = reverseLL(slow->next);
slow = slow->next;
while (slow){
if (head->val != slow->val) return false;
head = head->next;
slow = slow->next;
}
return true;
}
ListNode* reverseLL(ListNode *node){
ListNode *res = NULL;
while (node){
ListNode *next = node->next;
node->next = res;
res = node;
node = next;
}
return res;
}
};