# [LeetCode] Rotate List (Java)

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->NULL and k = 2,
return 4->5->1->2->3->NULL.

## Analysis

This problem is not so difficult. We can use two pointers, a fast pointer and a slow pointer, to achieve this. However, there are some corner cases in this problem which needs to be careful.

Take 1 -> 2 -> 3 -> 4 -> 5 as an example. Now k is 2. Our fast pointer will be pointing to 3, and the slow pointer will be pointing to 1, which is 2 steps slow. And it will be easy to transform the list to what we want if we have such pointers.

## Complexity

The complexity is O(n + k).