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.

翻转完毕之后，记得处理下头尾即可

代码：

class Solution {public:    ListNode *reverseBetween(ListNode *head, int m, int n) {        if(head==NULL) return head;        //加一个哨兵        ListNode* headTag=new ListNode(0);        headTag->next=head;        ListNode* prep=headTag;        ListNode* p=headTag;        ListNode* q=headTag;        int prepStep2=m-1;        int qSteps=n-m;        while (prepStep2>0)        {            prep=prep->next;            --prepStep2;        }        p=prep->next;        q=p->next;        ListNode* temp;        ListNode* tempFirst=p;        bool firstTag=true;        while (qSteps)        {            temp=q->next;            q->next=p;            if(firstTag) {p->next=NULL;firstTag=false;}            p=q;            q=temp;            --qSteps;        }        prep->next=p;        tempFirst->next=q;        return headTag->next;    }};