Given a list, rotate the list to the right by k places, where k is non-negative. Example: Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL.
Since n may be a large number compared to the length of list. So we need to know the length of linked list.After that, move the list after the (l-n%l )th node to the front to finish the rotation.
Ex: {1,2,3} k=2 Move the list after the 1st node to the front
Ex: {1,2,3} k=5, In this case Move the list after (3-5%3=1)st node to the front.
So the code has three parts.
Get the length
Move to the (l-n%l)th node
3)Do the rotation
public ListNode rotateRight(ListNode head, int n) {
if (head==null||head.next==null) return head;
ListNode dummy=new ListNode(0);
dummy.next=head;
ListNode fast=dummy,slow=dummy;
int i;
for (i=0;fast.next!=null;i++)//Get the total length
fast=fast.next;
for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
slow=slow.next;
fast.next=dummy.next; //Do the rotation
dummy.next=slow.next;
slow.next=null;
return dummy.next;
}
原文：http://www.cnblogs.com/apanda009/p/7898184.html