<p>Given the <code>head</code> of a singly linked list and two integers <code>left</code> and <code>right</code> where <code>left &lt;= right</code>, reverse the nodes of the list from position <code>left</code> to position <code>right</code>, and return <em>the reversed list</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev2ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], left = 2, right = 4
<strong>Output:</strong> [1,4,3,2,5]
</pre>

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

<pre>
<strong>Input:</strong> head = [5], left = 1, right = 1
<strong>Output:</strong> [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;= n &lt;= 500</code></li>
	<li><code>-500 &lt;= Node.val &lt;= 500</code></li>
	<li><code>1 &lt;= left &lt;= right &lt;= n</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you do it in one pass?
# Definition for singly-linked list.

class Solution:
    def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
        
        if left == right:
            return head
        
        cur = head
        cnt = 1
        
        prev = None
        while cnt < left:
            cnt += 1
            prev = cur
            cur = cur.next            

        continueNodeFromRight = cur 
        prev1 = None
        while cnt <= right:
            cnt += 1
            tmpNext = cur.next
            cur.next = prev1
            prev1 = cur
            cur = tmpNext

        continueNodeFromRight.next = cur
        if left == 1: 
            return prev1
        else:
            prev.next = prev1
            return head
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