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.

思路：

用四个指针分别记录第m-1，m， n， n+1个节点，然后把第m到n个节点之间的node的next指向前一个node。注意一些特殊情况的判断。

代码：

 1     ListNode *reverseBetween(ListNode *head, int m, int n) {
 2         // Note: The Solution object is instantiated only once and is reused by each test case.
 3         if(head == NULL)
 4             return NULL;
 5         if(m == n)
 6             return head;
 7         ListNode *msub1, *mp, *np, *nadd1, *last, *cur, *nextp, *tmp;
 8         if(m == 1)
 9             msub1 = NULL;
10         int index = 1;
11         tmp = head;
12         while(tmp){
13             if(index == m-1){
14                 msub1 = tmp;
15             }
16             if(index == m){
17                 mp = tmp;
18                 break;
19             }
20             tmp = tmp->next;
21             index++;
22         }
23         if(mp == NULL || mp->next == NULL)
24             return head;
25         last = mp;
26         cur = mp->next;
27         if(cur == NULL)
28             return head;
29         nextp = cur->next;
30         index = m+1;
31         while(index <= n){
32             if(index == n){
33                 np = cur;
34             }
35             cur->next = last;
36             last = cur;
37             cur = nextp;
38             if(nextp == NULL)
39                 break;
40             nextp = nextp->next;
41             index++;
42         }
43         nadd1 = cur;
44         if(msub1)
45             msub1->next = np;
46         else
47             head = np;
48         mp->next = nadd1;
49         return head;
50     }