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;

	}

巴特西

Rotate List ，反转链表的右k个元素

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