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.

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;

    }
};