题目

【判断回文链表】
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?

根据提交结果，这里还应该加上一个例子：
Input: 1->2->3->2->1
Output: true

思路

将链表的后半部分进行翻转，成为一个新链表，然后判断这两个链表的值是否相等，相等即为回文链表。
时间复杂度O(n)
由于只使用了三个指针，时间复杂度为O(1)

代码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) {
        ListNode fast = head, slow = head;
        while(fast != null && fast.next != null  ){
        	fast = fast.next.next;
        	slow = slow.next;

        }
        //特殊处理
        if(fast != null){
        	slow = slow.next;
        }
        //下面对后半段进行迭代法翻转
  
        ListNode newHead = null;
        
        
        while(slow != null){
        fast = slow.next; //把fast指向工作指针的下一个地址防止丢失
        slow.next = newHead; //指向新表头
        newHead = slow;
        slow = fast;       
        }

        //下面比较两个链表是否相同
        while(newHead != null){
        	if(head.val != newHead.val) return false;
        	head = head.next;
        	newHead = newHead.next;
        }
        return true;
    }
}

提交结果

Runtime: 1 ms, faster than 93.53% of Java online submissions for Palindrome Linked List.
Memory Usage: 42.6 MB, less than 31.50% of Java online submissions for Palindrome Linked List.