LeetCode(Rotate List) 旋转链表

题目要求：

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:

Given

1->2->3->4->5->NULL

and k =

2

,

return

4->5->1->2->3->NULL

.
思路： 首先求出链表的长度len， k = k % len，然后找到倒数第k个节点，添加一个虚拟头结点，然后将倒数第k节点到末尾节点用尾插法插入。

代码：

ListNode *rotateRight(ListNode *head, int k)
{
if(head == NULL || k <=0)
return head;
int len = ListLength(head);
k = k % len;
if(k == 0)
return head;
ListNode* first = head, *second = head;
for(size_t i = 0; i < k; ++i)
{
first = first->next;
}
while(first->next != NULL)
{
first = first->next;
second = second->next;
}
ListNode* rotate_head = second->next;
second->next = NULL;
ListNode* virtual_head = new ListNode(0);
virtual_head->next = head;
ListNode* cur = virtual_head;
while(rotate_head != NULL)
{
ListNode* node = rotate_head->next;
rotate_head->next = cur->next;
cur->next = rotate_head;
rotate_head = node;
cur = cur->next;
}
ListNode* ret = virtual_head->next;
delete virtual_head;
virtual_head = NULL;
return ret;
}

int ListLength(ListNode* head)
{
if(head == NULL)
return 0;
int len = 0;
while(head != NULL)
{
++len;
head = head->next;
}
return len;
}