Rotate List ，反转链表的右k个元素

问题描述：

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

.

算法分析：这个k是可以超过链表长度的，所以要对k取模。

public ListNode rotateRight(ListNode head, int k) {
int lenth = 0;//链表长度
ListNode headd = head;
while (headd != null) {
lenth++;
headd = headd.next;
}
if (head == null || head.next == null || k % lenth == 0) {
return head;
}

ListNode headc = head;
for (int i = 1; i < lenth - k % lenth; i++) {
headc = headc.next;
}
ListNode newHead = headc.next;
ListNode p = newHead;
while (p.next != null) {
p = p.next;
}
p.next = head;
headc.next = null;

return newHead;
}