[Leetcode] Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

 

no words!!!!

 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     ListNode *reverseBetween(ListNode *head, int m, int n) {
12         ListNode *pre, *last;
13         ListNode *h = new ListNode(-1);
14         h->next = head;
15         pre = h;  last = head;
16         for (int i = 1; i < m; ++i) {
17             pre = pre->next;
18         }
19         for (int i = 1; i < n; ++i) {
20             last = last->next;
21         }
22         if (pre->next == last) return h->next;
23         ListNode *pos = pre->next, *tmp;
24         pre->next = last;
25         pre = last->next;
26         while (pos != NULL && pos != last) {
27             tmp = pos->next;
28             pos->next = pre;
29             pre = pos;
30             pos = tmp;
31         }
32         pos->next = pre;
33         return h->next;
34     }
35 };

 

posted @ 2014-04-04 18:09  Eason Liu  阅读(203)  评论(0)  编辑  收藏  举报