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[yuseok89] WEEK 09 Solutions #2824

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22 changes: 22 additions & 0 deletions linked-list-cycle/yuseok89.py
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@dalestudy dalestudy Bot Aug 19, 2026

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🏷️ 알고리즘 패턴 분석

linked-list-cycle/yuseok89.py 
# TC: O(N)
# SC: O(1)
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        if not head or not head.next:
            return False

        slow, fast = head, head.next
        while slow != fast:
            if not fast or not fast.next:
                return False

            slow, fast = slow.next, fast.next.next

        return True
  • 패턴: Two Pointers, Fast & Slow Pointers
  • 설명: 해당 코드는 느린 포인터와 빠른 포인터를 이용해 순환 여부를 판별하는 전형적인 Fast & Slow Pointers 패턴이다. 두 포인터의 속도 차이를 이용해 사이클 존재 여부를 O(N) 시간, O(1) 공간으로 판단한다.

📊 시간/공간 복잡도 분석

복잡도
Time O(n)
Space O(1)

피드백: 빠른 포인터와 느린 포인터를 이용해 순회하므로 추가 메모리 없이 사이클 여부를 판단할 수 있다.

개선 제안: 현재 구현이 적절해 보입니다.

💡 풀이에 시간/공간 복잡도를 주석으로 남겨보세요!

@alphaorderly alphaorderly Aug 20, 2026

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조건문 부분에 같은지 비교를 하는것도 꽤 괜찮은 아이디어시네요!
다만 아주 개인적인 의견인데, slow와 fast를 비교하는것은 참조를 비교하는거라 is를 쓰는게 더 낫지 않을까? 하는 생각이 있었어요
좋은 코드 보고 갑니다!

Original file line number Diff line number Diff line change
@@ -0,0 +1,22 @@
# TC: O(N)
# SC: O(1)
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution:
def hasCycle(self, head: Optional[ListNode]) -> bool:
if not head or not head.next:
return False
Comment on lines +11 to +12

@DaleSeo DaleSeo Aug 20, 2026

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노드가 하나뿐이고 순환이 없는 입력이라면 fast가 None으로 시작하므로, 반복문 안의 if not fast 가 곧바로 False를 반환해 줄텐데요. 그렇다면 여기서 not head.next까지 확인하는 것은 어떤 경우를 대응하기 위한 목적인가요?


slow, fast = head, head.next
while slow != fast:
if not fast or not fast.next:
return False

slow, fast = slow.next, fast.next.next

return True

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