Reverse Linked List II

题目描述如下:

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.

可以看出编号是从1开始的，最讨厌这种不是从0开始的了，处理跑的长度的时候超恶心啊；

 #define ln ListNodeclass Solution {public:ln* reverseBetween(ln* head,int m,int n){    if ( !head || !head->next || m<1 )return head;    n=n-m+1;ln* guard=new ln(-1);guard->next=head;ln* pPre=guard;while(pPre&&--m){pPre=pPre->next;}//not enough m nodes;if ( pPre==NULL){delete guard;return head;}ln* pAfter=pPre->next;while(pAfter&&n--){pAfter=pAfter->next;}_reverseList(pPre,pAfter);ln* ans=guard->next;delete guard;return ans;}void _reverseList(ln*& pPre,ln* pAfter){ln* pCur=pPre->next;pPre->next=pAfter;while(pCur!=pAfter){ln* tmp=pCur->next;pCur->next=pPre->next;pPre->next=pCur;pCur=tmp;}}};