1，找到中间节点，并把前半部的链表以NULL结尾，
2，把后半部的链表反转，
3,前后合并两个链表

另外值得注意的是，对于中间节点的寻找，第二种方法中间总是中间位置偏后，第一种对于奇数个数的中间节点，偏前。

        fast = fast->next;
        if(fast){
            fast = fast->next;
            slow = slow->next;
        }
{1 2 3 4} slow = 3 
{1 2 3}  slow = 2

        fast = fast->next;
        if(fast){
            fast = fast->next;      
        }
        slow = slow->next;
{1 2 3 4} slow = 3 
{1 2 3}  slow = 3

struct ListNode * reverse(struct ListNode *head)
{
    if(head == NULL || head->next == NULL)
        return head;
    struct ListNode * prev, *curr, *next;
    curr = head;
    prev = next = NULL;

    while(curr){
        next = curr->next;
        curr->next = prev;
        prev =  curr;
        curr = next;
    }


    return prev;
}

void reorderList(struct ListNode* head) {

    if(head == NULL || head->next == NULL)
        return head;
    struct ListNode *fast, *slow;
    fast = head->next;
        slow = head;
    while(fast){
        fast = fast->next;
        if(fast){
            fast = fast->next;
            slow = slow->next;
        }

    }

    struct ListNode *tmp = slow;
    slow = slow->next;
    tmp->next = NULL;
    struct ListNode  *front = head;
    struct ListNode *back = reverse(slow);

    struct ListNode *dummy = calloc(1, sizeof(struct ListNode));
    struct ListNode *last = dummy;

    while(1){
        if(front&&back){
            last->next = front;
            front = front->next;
            last = last->next;

            last->next = back;
            back = back->next;
            last = last->next;
        }else if(front){
            last->next = front;
            front = front->next;
            last = last->next;

        }else if(back){
            last->next = back;
            back = back->next;
            last = last->next;
        }else{
            last->next = NULL;
            break;
        }
    }

    return dummy->next;
}