类似于找到链表的倒数第k个节点,然后把前后二段链表倒置前后顺序即可;
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(head==NULL)
return head;
ListNode* temp=head;
ListNode*last;
int len=0;
while(temp!=0)
{
len++;
if(temp->next==NULL)
last=temp;
temp=temp->next;
}
k%=len;
if(k==0)
return head;
int step=len-k-1;
ListNode* begin=head;
while(step>0)
{
begin=begin->next;
step--;
}
ListNode* ans=begin->next;
last->next=head;
begin->next=NULL;
return ans;
}
};