重排链表

将给定的单链表 L L： L_0→L_1→…→L_{n-1}→L_ nL0​→L1​→…→Ln−1​→Ln​
重新排序为：L_0→L_n →L_1→L_{n-1}→L_2→L_{n-2}→…L0​→Ln​→L1​→Ln−1​→L2​→Ln−2​→…
要求使用原地算法，不能改变节点内部的值，需要对实际的节点进行交换。
例如：
对于给定的单链表{10,20,30,40}，将其重新排序为{10,40,20,30}.

思路：

1，找到链表的中间节点

2，反转后面的链表

3，归并的方式合并前后链表

4，注意最后需要将前链表的最后一个节点的next改为null，否则会有环

代码

private static void reorderList(ListNode head) {
        if (head == null || head.next == null) {
            return;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        ListNode after = slow.next;
        ListNode pre = null;
        while (after != null) {
            ListNode temp = after.next;
            after.next = pre;
            pre = after;
            after = temp;
        }
        ListNode first = head;
        after = pre;
        while (after != null) {
            ListNode fTemp = first.next;
            ListNode aTemp = after.next;
            first.next = after;
            after.next = fTemp;


            first = fTemp;
            after = aTemp;


        }

        first.next = null;

        System.out.println(0);
    }