61 Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.
The Idea: Find the length of the list, and store the tail end of the list. Then mod k with the length of the list, because when k>length list, the rotation would cycle about itself, leaving a remainder behind. Then subtract k with the length. This will be the pivot point we want to get to from head. Point the tail end to the front of the list and have the pivot point to null.
Complexity: O(n) time and O(1) space
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def rotateRight(self, head, k):
:type head: ListNode
:type k: int
:rtype: ListNode
if not head:
return None
# get the length
length_list = 1
tail = head
while tail.next:
length_list += 1
tail = tail.next
# ineffective measures: if k perfectly divides the
# length, then the rotation would wrap about itself
remainder = k % length_list
if remainder == 0:
return head
go_right = length_list - remainder
iter_right = head
while go_right > 1:
go_right -= 1
iter_right = iter_right.next
tail.next = head
new_head = iter_right.next
iter_right.next = None
return new_head