<p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>

<p><code>k</code> is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of <code>k</code> then left-out nodes, in the end, should remain as it is.</p>

<p>You may not alter the values in the list&#39;s nodes, only nodes themselves may be changed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 2
<strong>Output:</strong> [2,1,4,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 3
<strong>Output:</strong> [3,2,1,4,5]
</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;= 5000</code></li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverse(self,head, start, end):
        # Ensure the 'end' is within the bounds of the list
        end = min(end, len(head) - 1)
        while start < end:
            head[start], head[end] = head[end], head[start]
            start += 1
            end -= 1

    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        l = []
        temp = head
        while temp!=None:
            l.append(temp.val)
            temp = temp.next

        for i in range(0, len(l), k):
            # Reverse the sublist from index i to i+k-1, but ensure the end does not go out of bounds
            if i + k - 1 < len(l):  # Only reverse if there are 'k' elements left
                self.reverse(l, i, i + k - 1)

        dummy = ListNode(-1)
        temp = dummy
        for i in l:
            temp.next = ListNode(i)
            temp = temp.next
        
        return dummy.next
17800