★leetcode★ reorder-list

题目：

Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-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}.

 

 

（这个题目没有头结点）

思路：

  1. 通过快慢指针找出中心点位置；
  2. 把链表分成两部分；
  3. 对后一半链表逆序；
  4. 那后一半链表插入到前一半中，得到结果 
     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     {
13         if(head == NULL || head->next == NULL || head->next->next == NULL)
14             return;
15         ListNode *slow, *fast, *mid;
16         slow = head;
17         fast = head->next;
18         while(fast != NULL && fast->next != NULL)
19         {
20             slow = slow->next;
21             fast = fast->next->next;
22         }
23 
24 
25         mid = slow;
26         ListNode *Back, *p;
27         Back = mid->next;//后半链表开头
28         mid->next = NULL;
29         p = Back->next;
30         Back->next = NULL;
31         //后半链表逆序，后半链表Back
32         while(p)
33         {
34             ListNode *tmp;
35             tmp = p->next;
36             p->next = Back;
37             Back = p;
38             p = tmp;
39         }
40 
41 
42 
43         ListNode *Front = head;
44         while(Front != NULL && Back != NULL)
45         {
46             ListNode *temp1, *temp2;
47             temp1 = Front->next;
48             temp2 = Back->next;
49             Front->next = Back;
50             Back->next = temp1;
51             Front = temp1;
52             Back = temp2;
53         }
54 
55     }
56 };

     

     

posted on 2017-06-22 17:01  xdzhanght  阅读(146)  评论(0)  编辑  收藏  举报

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