92. 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.

 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){
13             return head;
14         }
15         ListNode* newHead = new ListNode(-1);
16         newHead->next = head;
17         
18         
19         //因为m 可能为1,n为2,此时需要三个结点才能完成转换，
20         //所以pNode设为newHead，
21         //这样newHead结点,第一个，第二个结点就可以完成转换
22         ListNode* pNode = newHead;
23         
24         for(int i = 1; i < m ; i++){
25             pNode = pNode->next;
26         }
27         
28         ListNode* pm = pNode->next;;
29         
30         for(int i = m ; i < n; i++){
31             ListNode* n = pm->next;
32             pm->next = n->next;
33             n->next = pNode->next;
34             pNode->next = n;
35         }
36         return newHead->next;
37     }
38 };