README_EN.md

234. Palindrome Linked List

中文文档

Description

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?

Solutions

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def isPalindrome(self, head: ListNode) -> bool:
        if head is None or head.next is None:
            return True
        slow, fast = head, head.next
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
        pre, cur = None, slow.next
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        while pre:
            if pre.val != head.val:
                return False
            pre, head = pre.next, head.next
        return True

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) {
            return true;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null) {
            if (pre.val != head.val) {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

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