LeetCode Reorder List 

 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  
12     ListNode* reverse(ListNode* head) {
13         if (head == NULL) {
14             return NULL;
15         }
16         ListNode* pre = NULL;
17         ListNode* cur = head;
18         while (cur != NULL) {
19             ListNode* tmp = cur->next;
20             cur->next = pre;
21             pre = cur;
22             cur = tmp;
23         }
24         return pre;
25     }
26 
27 
28     void reorderList(ListNode *head) {
29         if (head == NULL || head->next == NULL) {
30             return;
31         }
32         int num = 0;
33         ListNode* cur = head;
34         while (cur != NULL) {
35             num++;
36             cur = cur->next;
37         }
38         // seperate list, find middle node as the head of the new list
39         int ridx = num / 2;
40         ListNode* rhead = NULL;
41         cur = head;
42         for (int i=0; true; i++) {
43             if (i == ridx - 1) {
44                 rhead = cur->next;
45                 cur->next = NULL;
46                 break;
47             }
48             cur = cur->next;
49         }
50  
51         rhead = reverse(rhead);
52         
53         cur = head;
54         head = head->next;
55     
56         cur->next = rhead;
57         cur = rhead;
58         rhead = rhead->next;
59             
60         while (head != NULL && rhead != NULL) {
61             cur->next = head;
62             cur = head;
63             head = head->next;
64                 
65             cur->next = rhead;
66             cur = rhead;
67             rhead = rhead->next;
68         }
69         cur->next = rhead; // there must be only one node left(odd case) or NULL(even case)
70     }
71 
72 };

这题跟Copy List with Random Pointer 那题类似在生成结果前都对原先的链表结构进行了调整，这里就是把链表的后半段给逆序了一下，这样我们就能轻松的取出节点对(0, n), (1, n-1)...了（序号是原先未修改时的，[0, n]）

 

第二轮：

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}.

 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  // 10:26
10 class Solution {
11 public:
12     void reorderList(ListNode *head) {
13         if (head == NULL) {
14             return;
15         }
16         ListNode fakeHead(0);
17         fakeHead.next = head;
18         ListNode* slow = &fakeHead;
19         ListNode* fast = &fakeHead;
20         while (fast != NULL && fast->next != NULL) {
21             slow = slow->next;
22             fast = fast->next->next;
23         }
24         ListNode* second = slow->next;
25         slow->next = NULL;
26         
27         ListNode* pre = NULL;
28         ListNode* cur = second;
29         while (cur != NULL) {
30             ListNode* tmp = cur->next;
31             cur->next = pre;
32             pre = cur;
33             cur = tmp;
34         }
35         ListNode* ahead = fakeHead.next;
36         ListNode* bhead = pre;
37         
38         fakeHead.next = NULL;
39         ListNode* last = &fakeHead;
40         
41         while (ahead != NULL && bhead != NULL) {
42             last->next = ahead;
43             last = ahead;
44             ahead = ahead->next;
45             last->next = bhead;
46             last = bhead;
47             bhead = bhead->next;
48         }
49         last->next = ahead;
50     }
51 };

 取中间点的方法改进了一下

 

posted @ 2014-03-15 10:35  卖程序的小歪  阅读(171)  评论(0)  编辑  收藏  举报