25. Reverse Nodes in k-Group

📋 Đề Bài

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Constraints:

The number of nodes in the list is n.
1 <= k <= n <= 5000
0 <= Node.val <= 1000

Follow-up: Can you solve the problem in O(1) extra memory space?

🧠 Thuật Toán & Kỹ Thuật

Linked List (Danh sách liên kết)

⏱️ Thời gian O(n²)

💾 Không gian O(n)

💻 Lời Giải

Python 0025-reverse-nodes-in-k-group.py

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        currNode = head
        listNode = []
        
        while currNode:
            listNode.append(currNode.val)
            currNode = currNode.next
            
        n = len(listNode)
        
        for i in range(0, n + 1, k):
            listNode[i - k:i] = listNode[i - k:i][::-1]
        
        answNode = ListNode(0)
        dummNode = answNode
        
        for i in range(n):
            dummNode.next = ListNode(listNode[i])
            dummNode = dummNode.next
            
        return answNode.next