<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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
 class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        if (!head || left == right) return head;
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* prev = dummy;
        for (int i = 1; i < left; ++i) prev = prev->next;
        ListNode* curr = prev->next;
        ListNode* next = nullptr;
        for (int i = 0; i < right - left; ++i) {
            next = curr->next;
            curr->next = next->next;
            next->next = prev->next;
            prev->next = next;
        }
        return dummy->next;
    }
};
10400