题目：LCR 027

解法一：快慢指针 + 反转链表

1. 取中间节点的下一节点，即为链表后半部分。当链表为(1,2,3,4,null)时，取节点3。当链表为(1,2,3,4,5,null)时，取节点4。
2. 反转后半部分链表
3. 前半链表与后半链表同时开始向后遍历，验证回文串
4. 恢复链表

    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) return true;
        ListNode slow = head, fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode reverse = reverseList(slow.next), node1 = head, node2 = reverse;
        while (node2 != null) {
            if (node2.val != node1.val) return false;
            node2 = node2.next;
            node1 = node1.next;
        }
        reverseList(reverse);
        return true;
    }
    public ListNode reverseList(ListNode head) {
        ListNode prev = null, curr = head, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;