Move the Tail to the Front

Medium

You are given the head of a singly linked list and an integer k. Do the following k times: detach the last node and put it at the front of the list. Return the head of the resulting list. k can be far larger than the list's length, so literally repeating the step k times will not do.

Example 1

head=list_node([10, 20, 30, 40]), k=1 → [40, 10, 20, 30]

The last node 40 moves to the front: [40, 10, 20, 30].

Example 2

head=list_node([5, 6]), k=3 → [6, 5]

Two steps bring the list back to where it was, so three steps are the same as one: [6, 5].

Example 3

head=list_node([]), k=7 → None

An empty list stays empty.

Constraints

0 ≤ number of nodes ≤ 500
-100 ≤ Node.val ≤ 100
0 ≤ k ≤ 2·10⁹

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Topics

Linked List Two Pointers

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Input

head=list_node([10, 20, 30, 40]), k=1

Expected

[40, 10, 20, 30]

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