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.

解题思路：

先找到需要翻转的起始节点，然后，翻转其后的n - m 个节点。 注意处理翻转的起始节点为head的情况。

代码：

 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         if (head == NULL) return NULL;

13

14         ListNode *pre_node = NULL, *mid_first = head;

15         for (int i = 1; i < m; i++) {

16             pre_node = mid_first;

17             mid_first = mid_first->next;

18         }

19         ListNode *last = mid_first->next, *pre = mid_first;

20         for (int i = m; i < n; i++) {

21             ListNode *tmp = last->next;

22             last->next = pre;

23             pre = last;

24             last = tmp;

25         }

26         if (pre_node == NULL) {

27             head = pre;

28             mid_first->next = last;

29         } else {

30             pre_node->next = pre;

31             mid_first->next = last;

32         }

33         return head;

34     }

35 };

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