Add reverse-linked-list sol

Author: shinsj4653

reverse-linked-list/shinsj4653.py

@@ -1,15 +1,85 @@
 """
 [문제풀이]
 # Inputs
-
+- head of singly linked list
 # Outputs
-
+- return the reversed list
 # Constraints
-
+- The number of nodes in the list is the range [0, 5000].
+- 5000 <= Node.val <= 5000
 # Ideas
+1. 스택 활용?
+첫 head의 노드부터 스택에 넣기
+맨 끝 도달하면
+스택 pop 하면서 새 리스트 만들기??
+
+마지막 요소를 인식 못하나?
 
 [회고]
 
 """
+# 첫 제출
+
+# 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]:
+
+        ret = ListNode()
+        ret_head = ret
+        st = []
+
+        while True:
+            # 왜 마지막 요소일 때 안나가지고 자꾸 NoneType object has no attribute 'val' 오류 뜸
+            st.append(head.val)
+            if head.next is None:
+                break
+            head = head.next
+
+        while st:
+            val = st.pop()
+            node = ListNode(val, None)
+            ret_head.next = node
+            ret_head = ret_head.next
+
+        return ret.next
+
+# gpt 답변
+
+## 스택 유지
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if not head:
+            return None
+
+        st = []
+        while head:
+            st.append(head.val)
+            head = head.next
+
+        dummy = ListNode(0)
+        current = dummy
+
+        while st:
+            current.next = ListNode(st.pop())
+            current = current.next
+
+        return dummy.next
+
+## 스택 없이 포인터를 뒤집는 방식
+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