Reverse Linked List II

Solution:

Very fast but terrible memory usage. Can be refactored for sure
Iterate through linked list extracting values
Iterate through values up until left. Append to output using queue
From left to right using insert(0) insert values into reverseVal array
Append values in this array to output (since we used insert(0) it will already be reversed)
Append output from right to end
Iterate through output generating a new Linked List and return the head of the list

class ListNode:
    def __init__(self, val=0, next=None):
      self.val = val
      self.next = next

    def __repr__(self):
      return str(self.val)

class Solution:
    def reverseBetween(self, head, left, right):
        values, reverseVal, output = [], [], []
        current = head
        dummy = final = ListNode(-1)

        if left == right:
            return head

        while current:
            values.append(current.val)
            current = current.next
        
        current = head

        for i in range(left-1):
            output.append(values[i])

        for i in range(left-1, right):
            reverseVal.insert(0, values[i])
        
        for i in range(len(reverseVal)):
            output.append(reverseVal)

        for i in range(right, len(values)):
            ouput.append(values[i])
        
        for i in range(len(output)):
            final.next = ListNode(output[i])

        final = final.next
    
    return dummy.next

head = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))
left = 2
right = 4

p1 = Solution()
print(p1.reverseBetween(head, left, right))