Reverse linked list ii

# 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]:
        dummy = ListNode()
        dummy.next = head
        curr = dummy
        for i in range(left - 1):
            curr = curr.next
        originalTail = curr
        reversedTail = curr.next
        prev = curr
        curr = curr.next
        for i in range(right - left + 1):
            next = curr.next
            curr.next = prev
            prev = curr
            curr = next

        originalTail.next = prev
        reversedTail.next = curr

        return dummy.next

Reverse Linked List II

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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

 

Follow up: Could you do it in one pass?