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;    }};