<p>You are given the <code>head</code> of a linked list, and an integer <code>k</code>.</p>

<p>Return <em>the head of the linked list after <strong>swapping</strong> the values of the </em><code>k<sup>th</sup></code> <em>node from the beginning and the </em><code>k<sup>th</sup></code> <em>node from the end (the list is <strong>1-indexed</strong>).</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/linked1.jpg" style="width: 400px; height: 112px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 2
<strong>Output:</strong> [1,4,3,2,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [7,9,6,6,7,8,3,0,9,5], k = 5
<strong>Output:</strong> [7,9,6,6,8,7,3,0,9,5]
</pre>

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

<ul>
	<li>The number of nodes in the list is <code>n</code>.</li>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</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 swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        stack = deque([])
        swap , ptr = None , head
        cnt = 0
        while ptr :
            if cnt < k :
                cnt += 1
            if cnt == k and not swap:
                swap = ptr
            stack.append(ptr)
            ptr = ptr.next
        ptr = None
        while cnt :
            ptr = stack.pop()
            cnt -= 1
        ptr.val , swap.val = swap.val , ptr.val
        return head
41000