LeetCode #61 Rotate List

Medium

Problem

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

Solution

Use two pointers runner and walker. Let runner runs entire list to calculate amount of nodes, say n, in list and stop at the last node. Then k % n = l for reducing unneeded move. At last, move walker to position n-l, which takes n-l-1 steps, and attach tail of list to head and reassign new head.

Complexity

runner running entire list takes O(n) time if n denotes to amount of nodes in this list. Then walker will take O(l) time. Because l = k % n, we know l will be less than n. Therefore, total time complexity is O(n+l) = O(n).

It’s trivial that it only takes O(1) extra space.