[leetcode-143-Reorder List]

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.
For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

思路很朴素：

1.用快慢指针将链表分为前后两部分。
2.将后部分链表用头插法翻转。
3.将翻转后的链表跟前半部分合并。

ListNode* reversal(ListNode* head)
    {//将链表用头插法 翻转
        if (head == NULL || head->next == NULL) return head;
        ListNode dump(0);        
        ListNode* next = head->next;
        while (next != NULL)
        {            
            head->next = dump.next;
            dump.next = head;
            head = next;
            next = next->next;
        }
        head->next = dump.next;
        dump.next = head;
        return dump.next;
    }
    ListNode* mergeList(ListNode* head, ListNode* head2)
    {//将list2 每一个元素插入到对应位置的list1元素的后面
        ListNode* temp = head;
        ListNode* p1 = head->next;
        ListNode* p2 = head2->next;
        while (head!=NULL && p1 != NULL && p2!= NULL)
        {
            head->next = head2;
            head2->next = p1;
            head = p1;
            p1 = p1->next;
            head2 = p2;
            p2 = p2->next;
        }
        head->next = head2;
        head2->next = p1;//处理最后一个结点
        return temp;
    }
    void reorderList(ListNode* head)
    {
        if (head == NULL || head->next == NULL || head->next->next == NULL) return;
        ListNode* slow = head, *fast = head;
        while (fast->next!=NULL && fast->next->next!=NULL)//当fast结点后只有一个或者没有结点时停止循环
        {
            slow = slow->next;
            fast = fast->next->next;
        }    
        ListNode*lastPart = reversal(slow->next); //后半部分链表 翻转        
        slow->next = NULL;//将链表分为前后两个部分    前一个链表长度大于或者等于后边链表长度

        head = mergeList(head, lastPart);
    }