Rotate List
最后更新于：2022-04-02 01:10:56
# 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).
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