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[freemjstudio] WEEK 08 Solutions #2806

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18 changes: 18 additions & 0 deletions reverse-linked-list/freemjstudio.py
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@dalestudy dalestudy Bot Aug 8, 2026

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🏷️ 알고리즘 패턴 분석

reverse-linked-list/freemjstudio.py 
# 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: Optional[ListNode]) -> Optional[ListNode]:
        prev= None
        current = head

        while current:
            next_node = current.next
            current.next = prev

            prev = current
            current = next_node

        return prev
  • 패턴: Two Pointers, Linked List
  • 설명: 리스트를 역순으로 뒤집는 문제로, 포인터를 서로 교차시키며 앞뒤 연결을 바꿔나가는 전형적인 Two Pointers 패턴에 해당합니다. 순회와 포인터 갱신을 통해 한 방향으로 진행합니다.

📊 시간/공간 복잡도 분석

복잡도
Time O(n)
Space O(1)

피드백: 현재 구현은 루프를 한 번 순회하며 각 노드의 next 포인터를 이전 노드로 바꿔 반전합니다.

개선 제안: 현재 구현이 적절해 보입니다.

💡 풀이에 시간/공간 복잡도를 주석으로 남겨보세요!

Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
# 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: Optional[ListNode]) -> Optional[ListNode]:
prev= None
current = head

while current:
next_node = current.next
current.next = prev

prev = current
current = next_node

return prev
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