错误解法一：正序查找的过程中，将前面的元素倒叙插入inverse链中，找到偶数中点时，对称查找。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public boolean isPalindrome(ListNode head) {if(head.next == null){// 只有一个元素return true;}ListNode inverse=null, find=head;while(find != null){// 把find倒叙插入inverse中ListNode temp=find.next;find.next = inverse;inverse = find;find = temp;while(find != null && inverse != null && find.val == inverse.val){// 找到中点在find和find.next之间find = find.next;inverse = inverse.next;}}if(find==null && inverse==null ){return true;}return false;}
}

错误原因：没有考虑奇数情况

解法一：将数据转存到列表中，用双指针比较。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public boolean isPalindrome(ListNode head) {List<Integer> list = new ArrayList<>();// 将所有数据放入list中while(head!=null){list.add(head.val);head=head.next;}// 使用左右指针查找int left=0,right=list.size()-1;while(left<right){if(list.get(left)!=list.get(right)){return false;}right--;left++;}return true;}
}