# 20. \[Linked List] Reverse Linked List II

92\. Reverse Linked List IIMedium9.3K420Companies

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*.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/19/rev2ex2.jpg)

<pre><code><strong>Input: head = [1,2,3,4,5], left = 2, right = 4
</strong><strong>Output: [1,4,3,2,5]
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: head = [5], left = 1, right = 1
</strong><strong>Output: [5]
</strong></code></pre>

**Constraints:**

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

**Follow up:** Could you do it in one pass?

{% embed url="<https://leetcode.com/problems/reverse-linked-list-ii/>" %}

> Time: O(N)

> Space: O(1)

```javascript
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} left
 * @param {number} right
 * @return {ListNode}
 */
var reverseBetween = function(head, left, right) {
    if (left === right) 
        return head;

    const dummy = new ListNode(-1);
    dummy.next = head;

    let start = head;
    let end = head;

    let curr = dummy;
    let cnt = 0;
    while (curr !== null) {
        if (cnt === left - 1) {
            start = curr;
        }

        if (cnt === right) {
            end = curr;
        }

        curr = curr.next;
        cnt++;
    }

    let h = start.next;
    let t = end.next;
    curr = h;
    let prev = t;
    while (curr !== null && curr !== t) {
        const next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }

    start.next = prev;


    return dummy.next;
};
```


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