<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 or not k:
            return head

        n = 0
        curr = head
        while curr:
            last = curr
            curr = curr.next
            
            n += 1
        
        k = k % n
        if k == 0:
            return head
        
        nodes_to_move = n - k

        curr = head
        for i in range(nodes_to_move - 1):
            curr = curr.next
        
        prev = curr
        new_head = prev.next
        prev.next = None
        last.next = head
        return new_head

        


        
























    def rotateRight2(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if not head:
            return None
        curr = head
        list_len = 0
        last_node = None
        while curr:
            if not curr.next:
                last_node = curr
            list_len += 1
            curr = curr.next

        adjusted_rotation = k % list_len
        if adjusted_rotation == 0:
            return head

        idx_new_head = list_len - adjusted_rotation

        curr = head
        for _ in range(idx_new_head - 1):
            curr = curr.next
        
        tail = curr
        last_node.next = head
        head = tail.next
        tail.next = None
        return head

16600