我们可以采用双指针的办法进行,如下图:

如果链表长度为偶数,则直接从第二个指针的位置开始对链表进行反转;如果是奇数,则从第二指针的下一位进行链表反转

代码实现:

    public static void main(String[] args) {
        ListNode next4 = new ListNode(1, null);
        ListNode next3 = new ListNode(2, null);
        ListNode next2 = new ListNode(2, next3);
        ListNode next1 = new ListNode(1, next2);
        System.out.println("是否是回文链表:" + isPalindrome(next1));
    }

    //回文链表
    private static boolean isPalindrome(ListNode node) {
        ListNode fast = node, slow = node;
        while (fast != null && fast.node != null) {
            fast = fast.node.node;
            slow = slow.node;
        }
        if (fast != null) {
            slow = slow.node;
        }
        slow = reverse(slow);
        while (slow != null) {
            if (slow.val != node.val) {
                return false;
            }
            slow = slow.node;
            node = node.node;
        }
        return true;
    }

    private static ListNode reverse(ListNode slow) {
        if (slow.node == null) {
            return slow;
        }
        ListNode temp = null,next;
        while (slow != null) {
            next = slow.node;
            slow.node = temp;
            temp = slow;
            slow = next;
        }
        return slow;
    }