#25

Reverse Nodes in K Group

hard· Linked Listruns: 0

Given the head of a linked list, reverse the nodes of the list k at a time and return the modified head. k is a positive integer no larger than the length of the list. If the number of nodes is not a multiple of k, the remaining nodes at the end stay as they are. You may not alter the values; only nodes themselves may be changed.

wpm 0acc 100%time 0:000 / 701

class Solution:
    def reverseKGroup(
        self, head: ListNode | None, k: int
    ) -> ListNode | None:
        dummy = ListNode(0, head)
        group_prev = dummy
        while True:
            kth = group_prev
            for _ in range(k):
                kth = kth.next
                if not kth:
                    return dummy.next
            group_next = kth.next
            prev = group_next
            curr = group_prev.next
            while curr != group_next:
                nxt = curr.next
                curr.next = prev
                prev = curr
                curr = nxt
            tmp = group_prev.next
            group_prev.next = kth
            group_prev = tmp

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