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 ListNode
class 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;
}
}
};