92. Reverse Linked List II

题目：

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.

直接上代码了，在讨论区学习的，自己没有思路。多看多学习吧。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n)
{
//模仿大神方法，自己没有想出来
if (m == n)   return head;
ListNode* dummy = new ListNode(0);
dummy->next = head;
ListNode* pre = dummy;
for (int i = 0; i < m - 1; i++)    pre =  pre->next;
ListNode* start = pre->next;
ListNode* next = start->next;
for (int i = 0; i < n - m; i++)
{
start->next = next->next;
next->next = pre->next;
pre->next = next;
next = start->next;
}
return dummy->next;
}
};