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.

A是A出来了，不过写的略挫，有点懒，不太想改了

 1 public class Solution {
 2     public ListNode reverseBetween(ListNode head, int m, int n) {      
 3         ListNode headNode = new ListNode(0);
 4         headNode.next = head;
 5         head = headNode;                            //插入一个头结点
 6         
 7         ListNode toEnd = head;
 8         ListNode toInsert = head;
 9         
10         for(int i = 1; i < m; i++){                        //找到插入点
11             toInsert = toInsert.next;
12             toEnd = toEnd.next;
13         }//for
14         if(m < n){                                        //第一个采用头结点插入法插入的
15             toEnd = toEnd.next;
16             ListNode temp = new ListNode(toEnd.val);
17             toInsert.next = temp;
18         }//if
19         
20         int count = n - m;
21         while(count > 0){                                //采用头插法
22             toEnd = toEnd.next;
23             ListNode temp = new ListNode(toEnd.val);
24             temp.next = toInsert.next;
25             toInsert.next = temp;
26             count--;
27         }//while
28         
29         if(m < n)
30         {
31             while(toInsert.next != null)
32                 toInsert = toInsert.next;
33             toInsert.next = toEnd.next;
34         }
35         
36         head = head.next;
37         return head;
38     }    
39 }