Q：将给定的单链表L： L 0→L 1→…→L n-1→L n,重新排序为： L 0→L n →L 1→L n-1→L 2→L n-2→…
要求使用原地算法，并且不改变节点的值
例如：对于给定的单链表{1,2,3,4}，将其重新排序为{1,4,2,3}.
A：
链表从中点分割成两个，后面的倒装后，再和前面的merge

    public static void reorderList(ListNode head) {
        if (head == null || head.next == null)
            return;
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        fast = slow.next;
        slow.next = null;
        ListNode mid = reverse(fast);
        merge(head, mid);
    }

    public static ListNode reverse(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode head0 = new ListNode(Integer.MIN_VALUE);
        ListNode node = head;
        ListNode node1;
        while (node != null) {
            node1 = node.next;
            node.next = head0.next;
            head0.next = node;
            node = node1;
        }
        return head0.next;
    }

    public static void merge(ListNode head, ListNode mid) {
        ListNode node = head;
        ListNode node1 = mid;
        ListNode node2;
        while(node!=null && node1!=null){
            node2 = node1.next;
            node1.next = node.next;
            node.next = node1;
            node = node1.next;
            node1 = node2;
        }
    }