<p>Given the <code>head</code> of a linked&nbsp;list, rotate the list to the right by <code>k</code> places.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/rotate1.jpg" style="width: 450px; height: 191px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 2
<strong>Output:</strong> [4,5,1,2,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/roate2.jpg" style="width: 305px; height: 350px;" />
<pre>
<strong>Input:</strong> head = [0,1,2], k = 4
<strong>Output:</strong> [2,0,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 500]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
	<li><code>0 &lt;= k &lt;= 2 * 10<sup>9</sup></code></li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if not head:
            return None
        dummy = ListNode(0, head)
        length = 0

        while dummy.next:
            dummy = dummy.next
            length += 1

        k = k % length
        if k == 0:
            return head

        predecessor = ListNode(0, head)
        while length-k > 0:
            predecessor = predecessor.next
            k += 1
        newHead = predecessor.next
        predecessor.next = None
        
        tmp = newHead
        while tmp.next:
            tmp = tmp.next
        tmp.next = head

        return newHead
16400