Leetcode#143 Reorder List

先把链表分割成前后两半，然后交叉融合

实践证明，凡是链表相关的题目，都应该当成工程类题目做，局部变量、功能函数什么的随便整，代码长了没关系，关键是清楚，不容易出错。

代码：

 1 ListNode *reverseList(ListNode *head) {
 2   if (!head) return head;
 3         
 4   ListNode *prev = head;
 5   head = head->next;
 6   prev->next = NULL;
 7   while (head) {
 8     ListNode *next = head->next;
 9     head->next = prev;
10     prev = head;
11     head = next;
12   }
13         
14   return prev;
15 }
16     
17 ListNode *mergeList(ListNode *head1, ListNode *head2) {
18   ListNode *next1, *next2;
19   ListNode *head = head1;
20         
21   while (head1 && head2) {
22     next1 = head1->next;
23     next2 = head2->next;
24     head1->next = head2;
25     if (next1)
26       head2->next = next1;
27     head1 = next1;
28     head2 = next2;
29   }
30         
31   return head;
32 }
33 
34 void reorderList(ListNode *head) {
35   if (!head)
36     return;
37             
38   ListNode *fast = head;
39   ListNode *slow = head;
40   while (fast && fast->next) {
41     fast = fast->next->next;
42     slow = slow->next;
43   }
44         
45   ListNode *head1 = head;
46   ListNode *head2 = reverseList(slow->next);
47   slow->next = NULL;
48   mergeList(head1, head2);
49 }