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.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        if (head.next == null || left == right) {
            return head;
        }
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy;
        for (int i = 0; i < left - 1; ++i) {
            pre = pre.next;
        }
        ListNode p = pre;
        ListNode q = pre.next;
        ListNode cur = q;
        for (int i = 0; i < right - left + 1; ++i) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        p.next = pre;
        q.next = cur;
        return dummy.next;
    }
}