Reverse Linked List II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0092.Reverse%20Linked%20List%20II/images/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?
Linked List
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseBetween(
        self, head: Optional[ListNode], left: int, right: int
    ) -> Optional[ListNode]:
        if head.next is None or left == right:
            return head
        dummy = ListNode(0, head)
        pre = dummy
        for _ in range(left - 1):
            pre = pre.next
        p, q = pre, pre.next
        cur = q
        for _ in range(right - left + 1):
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        p.next = pre
        q.next = cur
        return dummy.next
