From cd3d89ee64c06dcc999b6392d8c1db072d55475e Mon Sep 17 00:00:00 2001
From: socow <socowjh@gmail.com>
Date: Tue, 6 Jan 2026 18:05:33 +0900
Subject: [PATCH] Linked List Cycle

---
 linked-list-cycle/socow.py                    | 41 +++++++++++++++++++
 .../socow.py                                  | 17 --------
 2 files changed, 41 insertions(+), 17 deletions(-)
 create mode 100644 linked-list-cycle/socow.py

diff --git a/linked-list-cycle/socow.py b/linked-list-cycle/socow.py
new file mode 100644
index 0000000000..7a02e4c882
--- /dev/null
+++ b/linked-list-cycle/socow.py
@@ -0,0 +1,41 @@
+"""
+📚 141. Linked List Cycle
+
+📌 문제 요약
+- 연결 리스트에 사이클(순환)이 있는지 확인하기
+- 있으면 True, 없으면 False
+
+🎯 핵심 알고리즘
+- 패턴: 플로이드 순환 탐지 (토끼와 거북이)
+- 시간복잡도: O(n)
+- 공간복잡도: O(1)
+
+💡 핵심 아이디어
+1. slow는 한 칸씩, fast는 두 칸씩 이동
+2. 사이클이 있으면 → 둘이 언젠가 만남!
+3. 사이클이 없으면 → fast가 끝에 도달 (None)
+"""
+
+from typing import Optional
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def hasCycle(self, head: Optional[ListNode]) -> bool:
+        slow = head
+        fast = head
+        
+        while fast and fast.next:
+            slow = slow.next        # 거북이: 한 칸
+            fast = fast.next.next   # 토끼: 두 칸
+            
+            if slow == fast:        # 만났다!
+                return True
+        
+        return False  # fast가 끝에 도달 = 사이클 없음
+
diff --git a/longest-substring-without-repeating-characters/socow.py b/longest-substring-without-repeating-characters/socow.py
index 91baa03a38..63b8887f16 100644
--- a/longest-substring-without-repeating-characters/socow.py
+++ b/longest-substring-without-repeating-characters/socow.py
@@ -35,21 +35,4 @@ def lengthOfLongestSubstring(self, s: str) -> int:
         return max_len
 
 
-# Set을 사용한 방식 (더 직관적)
-class SolutionWithSet:
-    def lengthOfLongestSubstring(self, s: str) -> int:
-        char_set = set()
-        left = 0
-        max_len = 0
-        
-        for right in range(len(s)):
-            # 중복이 사라질 때까지 left 이동
-            while s[right] in char_set:
-                char_set.remove(s[left])
-                left += 1
-            
-            char_set.add(s[right])
-            max_len = max(max_len, right - left + 1)
-        
-        return max_len