# Rotate List ### Source - leetcode: [Rotate List | LeetCode OJ](https://leetcode.com/problems/rotate-list/) - lintcode: [(170) Rotate List](http://www.lintcode.com/en/problem/rotate-list/) ### Problem Given a list, rotate the list to the right by *k* places, where *k* is non-negative. #### Example Given `1->2->3->4->5` and k = `2`, return `4->5->1->2->3`. ### 题解 旋转链表，链表类问题通常需要找到需要处理节点处的前一个节点。因此我们只需要找到旋转节点和最后一个节点即可。需要注意的细节是 k 有可能比链表长度还要大，此时需要取模，另一个 corner case 则是链表长度和 k 等长。 ### Java ~~~ /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { /** * @param head: the List * @param k: rotate to the right k places * @return: the list after rotation */ public ListNode rotateRight(ListNode head, int k) { if (head == null) return head; ListNode fast = head, slow = head; int len = 1; for (len = 1; fast.next != null && len <= k; len++) { fast = fast.next; } // k mod len if k > len if (len <= k) { k = k % len; fast = head; for (int i = 0; i < k; i++) { fast = fast.next; } } // forward slow and fast while (fast.next != null) { fast = fast.next; slow = slow.next; } // return new head fast.next = head; head = slow.next; slow.next = null; return head; } } ~~~ ### 源码分析 由于需要处理的是节点的前一个节点，故最终的`while` 循环使用`fast.next != null`. k 与链表等长时包含在`len <= k`中。 ### 复杂度分析 时间复杂度 O(n)O(n)O(n), 空间复杂度 O(1)O(1)O(1).