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

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {      
        ListNode headNode = new ListNode(0);
        headNode.next = head;
        head = headNode;                            //插入一个头结点
        
        ListNode toEnd = head;
        ListNode toInsert = head;
        
        for(int i = 1; i < m; i++){                        //找到插入点
            toInsert = toInsert.next;
            toEnd = toEnd.next;
        }//for
        if(m < n){                                        //第一个采用头结点插入法插入的
            toEnd = toEnd.next;
            ListNode temp = new ListNode(toEnd.val);
            toInsert.next = temp;
        }//if
        
        int count = n - m;
        while(count > 0){                                //采用头插法
            toEnd = toEnd.next;
            ListNode temp = new ListNode(toEnd.val);
            temp.next = toInsert.next;
            toInsert.next = temp;
            count--;
        }//while
        
        if(m < n)
        {
            while(toInsert.next != null)
                toInsert = toInsert.next;
            toInsert.next = toEnd.next;
        }
        
        head = head.next;
        return head;
    }    
}