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Use an iterative approach to reverse each group of k nodes separately. This approach leverages an auxiliary dummy node and previous and current pointers to manage the connections at each step. We traverse the list, checking for complete sets of k nodes to reverse them. If a group of nodes has less than k nodes, leave them as they are.
Time Complexity: O(n), where n is the number of nodes in the list, as each node is processed once.
Space Complexity: O(1), since no extra space is used apart from pointers.
We count length of the list to check if a full k group exists. A dummy node facilitates manipulation of edge connections. For each complete k-length, we perform a series of node swaps. We adjust prev, current, and next pointers accordingly to achieve the reversal.
Recurse through the linked list, reversing k nodes at a time. If a full group of k nodes is found, reverse and connect through recursive calls. Terminate when fewer than k nodes are left. This approach inherently uses the call stack for management of reverse sequences.
Time Complexity: O(n), since we process each node once even recursively.
Space Complexity: O(n/k), the recursion depth in worst case equals the number of complete k-sized groups.
1class ListNode:
2 def __init__(self, val=0, next=None):
3 self.val = val
4 self.next = next
5
6class Solution:
7 def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
8 count = 0
9 current = head
10 while current and count < k:
11 current = current.next
12 count += 1
13 if count == k:
14 current = self.reverseKGroup(current, k)
15 while count > 0:
16 tmp = head.next
17 head.next = current
18 current = head
19 head = tmp
20 count -= 1
21 head = current
22 return head
23The function probes node groups, inspecting k node availability. Featuring recursive return principles, node alignment reverses within recursive function sweeps which link completed segments reverting at recursion levels until resolution.
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