[leetcode] 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.

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		if(m>=n)
			return head;
		ListNode *L=new ListNode(0);
		L->next=head;
		int count=1;
		ListNode *p,*tear,*cur,*next;
		p=L,cur=L->next; 
		for( ; cur ; cur=cur->next,count++,p=p->next ){
			if(count==m)
				break;
		}
		if(!cur)
			return NULL;
		tear=cur;
		next=cur->next;
		for( ; cur ; cur=next , count++){
			if(count>n)
				break;
    		next=cur->next;
			cur->next=p->next;
			p->next=cur;
		}
		tear->next=next;
		p=L->next;
		delete L;
		return p;
	}
};