[LeetCode]Reverse Linked List II

题目描述：(链接)

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.

解题思路:

头插法

 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 dummy(-1);
13         dummy.next = head;
14         ListNode *prev = &dummy;
15         for (int i = 0; i < m - 1; ++i) {
16             prev = prev->next;
17         }
18         
19         ListNode *head2 = prev;
20         prev = head2->next;
21         ListNode *current = prev->next;
22         for (int i = m; i < n; ++i) {
23             prev->next = current->next;
24             current->next = head2->next;
25             head2->next = current;
26             current = prev->next;
27         }
28         
29         return dummy.next;
30     }
31 };