[LeetCode] 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：找到中间节点，将链表的后半部分就地逆置。然后插入到前半部分。

　　　时间复杂度O(n),空间复杂度O(1) 1 /**

 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void reorderList(ListNode *head) {
12         if (head == NULL || head->next == NULL || head->next->next == NULL) 
13             return;
14         
15         int length = 0;
16         for (ListNode *pter = head; pter != NULL; pter = pter->next) {
17             ++length;
18         }
19         
20         int half_length = 1;
21         ListNode *half = head;
22         while (half_length < (length / 2)) {
23             ++half_length;
24             half = half->next;
25         }
26         
27        //reverse
28        ListNode *pter = half->next;
29        half->next = NULL;
30        while (pter != NULL) {
31            ListNode *next_node = pter->next;
32            pter->next = half->next;
33            half->next = pter;
34            pter = next_node;
35        }
36       

37       ListNode *pter2 = head;
38       pter = half->next;
39       while (pter2 != half) {
40           ListNode *next_node = pter->next;
41           half->next =next_node;
42           pter->next = pter2->next;
43           pter2->next = pter;
44           pter2 = pter2->next->next;
45           pter = next_node;
46       }
47     }
48 };