Reverse Nodes in k-Group

Problem statement

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3Output: [3,2,1,4,5]

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

Follow-up: Can you solve the problem in O(1) extra memory space?

My solution

/**
 * 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} k
 * @return {ListNode}
 */
var reverseKGroup = function(head, k) {
    let newHead;
    let tail;
    let temp = []
    
    while (head) {
        temp.push(head)
        head = head.next;
        
        if (temp.length === k) {
            let tempHead = temp[k-1];
            let tempTail = undefined;
            for (let x = k-2; x>=0; x--) {
                if (tempTail) {
                    tempTail.next = temp[x];
                    tempTail = tempTail.next;
                } else {
                    tempHead.next = temp[x];
                    tempTail = tempHead.next;
                }
            }
            
            if (newHead) {
                if (tail) tail.next = tempHead;
                tail = tempTail;
            } else {
                newHead = tempHead;
                tail = tempTail;
            }
            
            temp = []
        }
        
    }
    
    if (temp.length > 0) {
        for (let node of temp) {
            if (newHead) {
                if (tail) tail.next = node;
                tail = node;
            } else {
                newHead = node;
            }
        }
    }
    
    if (tail) tail.next = null;
    
    return newHead
};