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.
ref: http://fisherlei.blogspot.com/2013/01/leetcode-rotate-list.html
首先从head开始跑，直到最后一个节点，这时可以得出链表长度len。然后将尾指针指向头指针，将整个圈连起来，接着往前跑len – k%len，从这里断开，就是要求的结果了。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode rotateRight(ListNode head, int n) { if(head == null || n == 0) return head; ListNode p = head; int len = 1; while(p.next != null){ len++; p = p.next; } p.next = head; int steps = len - n%len; while(steps > 0){ p = p.next; steps--; } head = p.next; p.next = null; return head; } }
原文：http://www.cnblogs.com/RazerLu/p/3549268.html