# [61. Rotate List](https://leetcode.com/problems/rotate-list/description/)

## 题目

Given a linked list, rotate the list to the right by k places, where k is non-negative.

**Example 1**:  

```

Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL 

```

**Example 2**:  

```

Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL 

```

## 题目大意

旋转链表 K 次。


## 解题思路

这道题需要注意的点是，K 可能很大，K = 2000000000 ，如果是循环肯定会超时。应该找出 O(n) 的复杂度的算法才行。由于是循环旋转，最终状态其实是确定的，利用链表的长度取余可以得到链表的最终旋转结果。

这道题也不能用递归，递归解法会超时。

## 代码

```go

package leetcode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func rotateRight(head *ListNode, k int) *ListNode {
	if head == nil || head.Next == nil || k == 0 {
		return head
	}
	newHead := &ListNode{Val: 0, Next: head}
	len := 0
	cur := newHead
	for cur.Next != nil {
		len++
		cur = cur.Next
	}
	if (k % len) == 0 {
		return head
	}
	cur.Next = head
	cur = newHead
	for i := len - k%len; i > 0; i-- {
		cur = cur.Next
	}
	res := &ListNode{Val: 0, Next: cur.Next}
	cur.Next = nil
	return res.Next
}

```