Reverse Nodes in k-Groups in a List

ReverseNodesInKGroups.py

# Leetcode 25
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode):
        prev = None
        curr = head
        while curr:
            nxt = curr.next
            curr.next = prev
            prev = curr
            curr = nxt
        return prev, head  # prev = new head, head = new tail

    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:

        # reserving only required if k is more than 1
        if k < 2:
            return head

        # Make a dummy node so we can use before pointer
        dummy = ListNode(-1)
        dummy.next = head
        before = dummy

        # Exhaust all the nodes
        curr = head
        while curr:

            # Find the end of the list to be reversed
            K = k
            end = before
            while K and end.next:
                end = end.next
                K -= 1

            # if K nodes are not available break
            if K:
                break

            # rest of the list is next to end of the list to be reversed
            after = end.next

            # Isolate the group to be reversed
            before.next = None
            end.next = None

            # Reverse the group
            start, end = self.reverseList(curr)

            # Connect with before and after
            before.next = start
            end.next = after

            # Move to next group
            before = end
            curr = after

        return dummy.next