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->nullptr and k = 2, return 4->5->1->2->3->nullptr.
分析
先遍历一遍，得出链表长度 len，注意 k 可能大于 len，因此令 k% = len。将尾节点 next 指针
指向首节点，形成一个环，接着往后跑 len ? k 步，从这里断开，就是要求的结果了。
代码
1 public static ListNode rotateRight(ListNode head, int n) { 2 if (head == null || head.next == null || n == 0) 3 return head; 4 ListNode fast = head, slow = head, countlen = head; 5 ListNode newhead = new ListNode(-1); 6 int len = 0; 7 8 while (countlen != null) { 9 len++; 10 countlen = countlen.next; 11 } 12 13 n = n % len; 14 if (n == 0) 15 return head; 16 for (int i = 0; i < n; i++) { 17 fast = fast.next; // slow与fast间隔n+1 18 } 19 while (fast.next != null) { 20 slow = slow.next; // slow指向要倒着数的开始点的前一个位置。 21 fast = fast.next; 22 23 } 24 25 newhead = slow.next; 26 fast.next = head; 27 slow.next = null; 28 29 return newhead; 30 }
原文：https://www.cnblogs.com/ncznx/p/9160842.html