Problem URL: Reverse linked list II

First if the left and right position of the list is same, we can just return the list. Otherwise, we will take a pointer, traverse till the left position, then reverse the list in place till the right position. Once the reversal of the list is done, we will return the head of the list.

# 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]:
        if left == right:
            return head

        p = dummy = ListNode(0)
        dummy.next = head
        for _ in range(left - 1):
            p = p.next

        cur = p.next
        pre = None
        for _ in range(right - left + 1):
            cur.next, pre, cur = pre, cur, cur.next

        p.next.next = cur
        p.next = pre

        return dummy.next

Time Complexity: O(n)
Space Complexity: O(1)