LeetCode(Reverse Linked List II) 反转指定位置之间的节点

题目要求：

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.
代码：

ListNode *reverseBetween(ListNode *head, int m, int n) {
if(head == NULL)
return NULL;
if(m > n)
return head;

//创建一个虚拟节点，方便全部反转
ListNode* virtual_head = new ListNode(-1);
virtual_head->next = head;
ListNode* pre = virtual_head;//保存m位置之前的节点
int cnt = 1;
while(cnt++ < m && pre != NULL)//找到m位置之前的节点
pre = pre->next;
ListNode* tail = pre;
ListNode* rtail = pre->next;
ListNode* cur = NULL;
cnt = 1;
while(cnt++ < (n - m + 1))//用头插法进行反转
{
if(rtail->next == NULL)
break;
cur = rtail->next;
rtail->next = cur->next;
cur->next = tail->next;
tail->next = cur;
}
head = virtual_head->next;
delete virtual_head;
virtual_head = NULL;
return head;
}