LeetCode OJ：Reorder List（重序链表）

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}.

如上例子所示的重序链表问题，先找到中间节点，然后将链表分成两段。将第二段反转后依次插入第一段中就得到了完整的链表，代码如下所示：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     void reorderList(ListNode* head) {
         if(!head || !head->next) return;
         ListNode * fastNode = head, * slowNode = head;
         while(fastNode->next){
             fastNode = fastNode->next;
             if(fastNode->next){
                 slowNode = slowNode->next;
                 fastNode = fastNode->next;
             }
         }
         ListNode * p1 = head;
         ListNode * p2 = slowNode->next;
         slowNode->next = NULL;//将前一段链表的最后一个节点的下一个赋为值NULL
         //将第二个节点指向的链表颠倒过来
         ListNode * prev = NULL;
         ListNode * curr = p2;
         ListNode * tmpNode = NULL;
         while(curr){
             tmpNode = curr->next;
             curr->next = prev;
             prev = curr;
             curr = tmpNode;
         }
         p2 = prev;//颠倒之后的首节点
         ListNode * tmpNode1, * tmpNode2;
         while(p2){
             tmpNode1 = p1->next;
             p1->next = p2;
             tmpNode2 = p2->next;
             p2->next = tmpNode1;
             p1 = tmpNode1;
             p2 = tmpNode2;
         }
     }
 };

代码重复有点多，写的比较乱，见谅见谅。