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81 lines
1.6 KiB
Markdown
81 lines
1.6 KiB
Markdown
# [61. Rotate List](https://leetcode.com/problems/rotate-list/description/)
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## 题目
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Given a linked list, rotate the list to the right by k places, where k is non-negative.
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**Example 1**:
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```
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Input: 1->2->3->4->5->NULL, k = 2
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Output: 4->5->1->2->3->NULL
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Explanation:
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rotate 1 steps to the right: 5->1->2->3->4->NULL
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rotate 2 steps to the right: 4->5->1->2->3->NULL
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```
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**Example 2**:
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```
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Input: 0->1->2->NULL, k = 4
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Output: 2->0->1->NULL
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Explanation:
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rotate 1 steps to the right: 2->0->1->NULL
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rotate 2 steps to the right: 1->2->0->NULL
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rotate 3 steps to the right: 0->1->2->NULL
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rotate 4 steps to the right: 2->0->1->NULL
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```
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## 题目大意
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旋转链表 K 次。
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## 解题思路
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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。
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这道题也不能用递归，递归解法会超时。
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## 代码
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```go
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package leetcode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func rotateRight(head *ListNode, k int) *ListNode {
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if head == nil || head.Next == nil || k == 0 {
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return head
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}
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newHead := &ListNode{Val: 0, Next: head}
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len := 0
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cur := newHead
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for cur.Next != nil {
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len++
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cur = cur.Next
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}
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if (k % len) == 0 {
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return head
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}
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cur.Next = head
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cur = newHead
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for i := len - k%len; i > 0; i-- {
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cur = cur.Next
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}
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res := &ListNode{Val: 0, Next: cur.Next}
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cur.Next = nil
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return res.Next
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}
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``` |