Reverse Linked List II

Could you do it in one pass?

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:

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

# 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: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
      
      dummyHead = ListNode(0, next = head)
      
      #head = [1,2,3,4,5], left = 2, right = 4
      start = dummyHead
      cur = start.next
      
      #find the node right before "left"
      for _ in range(left - 1):
        start = start.next
        cur = start.next
      #start = the node right before "left"
#rever the needed
      
      prev = None
      for _ in range(right - left + 1):
        nxt = cur.next
        cur.next = prev         
        prev = cur
        cur = nxt
      # 4 > 3 > 2 > None
      
      #end = the node after "right"
      end = cur
#start.next = 2
      start.next.next = end
      # 4 > 3 > 2 > 5
      
      #cur = 4
      #connect start with 4
      start.next = prev
      # 1 > 4 > 3 > 2 > 5
      
      return dummyHead.next