234. Palindrome Linked List【Easy】【判断链表是否回文】

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?

/**
 * 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;
        //根据快慢指针求中点
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        if(fast != null) 
            slow = slow.next;
        //slow是中点，将中点反转后
        slow = reverse(slow);
        fast = head;
        while(slow != null) {
             if(slow.val != fast.val) 
                return false;
            slow = slow.next;
            fast = fast.next;
           
        }
        return true;
    }
    public ListNode reverse(ListNode head) {
        ListNode pre = null;
        ListNode next = null;
        while(head != null) {
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}