[KwonNayeon] Week 12 solutions

reorder-list/KwonNayeon.py

@@ -21,7 +21,7 @@
 - 포인터들을 번갈아가며 연결함
 
 노트:
-- 포인터 조작(연결리스트 뒤집기, 병합) 방법을 다 까먹어서 복습용 예시 주석을 추가해둠
+- 포인터 조작(연결리스트 뒤집기, 병합)이 어려움. 스택 풀이도 준비해두기.
 """
 # Definition for singly-linked list.
 # class ListNode:
@@ -31,60 +31,38 @@
 class Solution:
     def reorderList(self, head: Optional[ListNode]) -> None:
         # 초기 상태: 1->2->3->4->5
-
+        """
+        Do not return anything, modify head in-place instead.
+        """
         # 1단계: 중간 지점 찾기
-        slow = head         # slow = 1
-        fast = head         # fast = 1
+        slow = head
+        fast = head
+
         while fast and fast.next:
-            slow = slow.next    # slow: 1->2->3
-            fast = fast.next.next   # fast: 1->3->5->None
-        # 결과: slow는 3에 위치
-
+            slow = slow.next
+            fast = fast.next.next
+
         # 2단계: 뒷부분 뒤집기
-        prev = None         # prev = None
-        curr = slow.next    # curr = 4 (뒷부분 시작점)
+        prev = None
+        curr = slow.next
         slow.next = None    # 분리: 1->2->3  |  4->5
-        
+
         while curr:
-            # 1회전: curr=4, prev=None
-            next_temp = curr.next   # next_temp = 5
-            curr.next = prev        # 4->None
-            prev = curr             # prev = 4
-            curr = next_temp        # curr = 5
-            # 상태: 1->2->3  |  4->None, curr=5
-
-            # 2회전: curr=5, prev=4
-            next_temp = curr.next   # next_temp = None
-            curr.next = prev        # 5->4
-            prev = curr             # prev = 5
-            curr = next_temp        # curr = None (종료)
-            # 상태: 1->2->3  |  5->4->None
-
+            next_temp = curr.next
+            curr.next = prev
+            prev = curr
+            curr = next_temp
+
         # 3단계: 앞부분과 뒷부분 합치기
-        first = head        # first = 1->2->3
-        second = prev       # second = 5->4
-        
+        first = head
+        second = prev
+
         while second:
-            # 1회전: first=1, second=5
-            temp1 = first.next      # temp1 = 2
-            temp2 = second.next     # temp2 = 4
-
-            first.next = second     # 1->5
-            second.next = temp1     # 5->2
-            # 현재 상태: 1->5->2->3, 남은 second = 4
-
-            first = temp1           # first = 2
-            second = temp2          # second = 4
-
-            # 2회전: first=2, second=4
-            temp1 = first.next      # temp1 = 3
-            temp2 = second.next     # temp2 = None
-
-            first.next = second     # 2->4
-            second.next = temp1     # 4->3
-            # 현재 상태: 1->5->2->4->3
-
-            first = temp1           # first = 3
-            second = temp2          # second = None (종료)
-
-        # 최종 결과: 1->5->2->4->3
+            temp1 = first.next
+            temp2 = second.next
+
+            first.next = second
+            second.next = temp1
+
+            first = temp1
+            second = temp2

reverse-linked-list/KwonNayeon.py

@@ -39,7 +39,8 @@ def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
         prev = None
         curr = head
 
-        while curr is not None:
+        # 마지막 노드도 처리해야 함
+        while curr:
 
             temp = curr.next
             curr.next = prev