leetcode 91. 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.

[Solution]

 1 ListNode *reverseBetween(ListNode *head, int m, int n)
 2     {
 3         ListNode dummy = ListNode(INT_MIN), *p, *q, *r;
 4         dummy.next = head;
 5         if (head == NULL)
 6             return NULL;
 7
 8         p = q = &dummy;
 9         for (int i = 1; i <= n - m; i++)
10             p = p->next;
11         for (int i = 1; i < m; i++)
12         {
13             p = p->next;
14             q = q->next;
15         }
16         p = p->next;
17         r = p->next;
18         p->next = NULL;
19
20         p = q->next;
21         q->next = reverse(p);
22         p->next = r;
23
24         return dummy.next;
25     }
26
27     ListNode *reverse(ListNode *head)
28     {
29         ListNode *p = NULL, *q, *r;
30
31         q = head;
32         if (q == NULL)
33             return NULL;
34         r = q->next;
35         while (r != NULL)
36         {
37             q->next = p;
38             p = q;
39             q = r;
40             r = r->next;
41         }
42         q->next = p;
43
44         return q;
45     }