---
title: '92. Reverse Linked List II'
description: Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list
icon: dot
topics:
  - { name: "Linked List", slug: "linked-list" }
issue: "https://github.com/prdlk/leetcode/issues/83"
sidebar:
  label: 'Reverse Linked List II'
  badge: 'Medium'
---

::::warning
Could you do it in one pass?
::::

### Example 1:
- Input: `head = [1,2,3,4,5], left = 2, right = 4`
- Output: `[1,4,3,2,5]`

### Example 2:
- Input: `head = [5], left = 1, right = 1`
- Output: `[5]`

### Constraints:

- The number of nodes in the list is n.
- `1 <= n <= 500`
- `-500 <= Node.val <= 500`
- `1 <= left <= right <= n`

## Solution

```py
# 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: ListNode | None, left: int, right: int
    ) -> ListNode | None:
        cur, prev = head, None
        while left > 1:
            prev = cur
            cur = cur.next
            left, right = left - 1, right - 1

        tail, con = cur, prev
        while right:
            third = cur.next
            cur.next = prev
            prev = cur
            cur = third
            right -= 1

        if con:
            con.next = prev
        else:
            head = prev

        tail.next = cur
        return head
```