思路：将链表分成两段，然后将后面一段进行翻转，在逐项进行比较。

class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null){
            return true;
        }
        ListNode temp = head;
        ListNode temp2 = head;
        //只有两个数字时，后面一段的链表是第二个
        ListNode temp1 = head.next;
        int len = 0;
        //遍历出链表有效数字
        while(temp!=null){
            temp = temp.next;
            len++;
        }
        for(int i=0; i<len/2-1;i++){
            temp2 = temp2.next;
            temp1 = temp1.next;
        }
        //分成两部分
        temp2.next = null;
        //有效数字是奇数个，后一段的链表在往后移一位
        if(len%2!=0){
            temp1 = temp1.next;
        }
        return isEqual(head,reverse(temp1));
    }
    //翻转链表
    private ListNode reverse(ListNode head) {
        ListNode newHead = null;
        while (head != null) {
            ListNode nextNode = head.next;
            head.next = newHead;
            newHead = head;
            head = nextNode;
        }
        return newHead;
    }
	//逐个判断是否相等
    private boolean isEqual(ListNode l1, ListNode l2) {
        while (l1 != null && l2 != null) {
            if (l1.val != l2.val) return false;
            l1 = l1.next;
            l2 = l2.next;
        }
        return true;

    }
}