234. Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

Follow up:
Could you do it in O(n) time and O(1) space?

Solution

Approach 1: Two pointers.

Code (Python)

Approach 1:

class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        if not head or not head.next:
            return True
        slow, fast = head, head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        return self._have_same_prefix(head, self._reverse_list(slow.next))
    
    def _reverse_list(self, head):
        if not head or not head.next:
            return head
        new_head = self._reverse_list(head.next)
        head.next.next = head
        head.next = None
        return new_head
    
    def _have_same_prefix(self, head1, head2):
        node1, node2 = head1, head2
        while node1 and node2:
            if node1.val != node2.val:
                return False
            node1, node2 = node1.next, node2.next
        return True

Code (C++)

Approach 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode *slow = head;
        ListNode *fast = head;
        ListNode *slowPrev = NULL;
        while (fast && fast->next)
        {
            fast = fast->next->next; // need to be here before "slow" reverses the link.
            ListNode *slowNext = slow->next;
            slow->next = slowPrev;
            slowPrev = slow;
            slow = slowNext;
        }
        ListNode *p1 = slowPrev;
        ListNode *p2 = (fast == NULL) ? slow : slow->next;
        while (p1 && p2)
        {
            if (p1->val != p2->val)
                return false;
            p1 = p1->next;
            p2 = p2->next;
        }
        return true;
    }
};