> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/61-rotate-list.md).

# 61 Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

**Example:**

```
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

```python
# 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
```


---

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