234. Palindrome Linked List - Easy

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?

找中点，反转后半部分，再一一比较

time: O(n), space: O(1)

/*** 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 fast = head, slow = head;while(fast.next != null && fast.next.next != null) {fast = fast.next.next;slow = slow.next;}ListNode p1 = head, p2 = reverse(slow.next);slow.next = null;while(p1 != null && p2 != null) {if(p1.val != p2.val) {return false;}p1 = p1.next;p2 = p2.next;}return true;}private ListNode reverse(ListNode head) {if(head == null || head.next == null) {return head;}ListNode prev = null, cur = head;while(cur != null) {ListNode next = cur.next;cur.next = prev;prev = cur;cur = next;}return prev;}
}