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.

处理这个问题还是挺复杂的，需要考虑很多边界的测试用例。我总体的思路是先用循环标记m前一个节点和n后边一个节点，把n后边的节点首先作为当前逆转节点的pre，然后循环n-m次完成所选节点部分的逆序，然后将标记的m节点前一个节点指向逆序后部分的头节点即可。要考虑各种特殊情况，另外考虑即可。code如下：

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if(head==NULL || m<0 || n<0)
            return head;
        if(head->next == NULL || m==n)
            return head;
        ListNode *head2=NULL,*pre,*cur,*temp=head;
        for(int i=0; i
  
   next;
        }
        for(int i=m;i
   
    next; cur->next = pre; pre = cur; cur = temp; } if(m==1) return pre; head2->next = pre; return head; } };