[LeetCode Sync] Runtime - 0 ms (100.00%), Memory - 18.8 MB (38.13%)

Author: NinePiece2

Solutions/0206-reverse-linked-list/README.md

@@ -0,0 +1,34 @@
+<p>Given the <code>head</code> of a singly linked list, reverse the list, and return <em>the reversed list</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>Input:</strong> head = [1,2,3,4,5]
+<strong>Output:</strong> [5,4,3,2,1]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg" style="width: 182px; height: 222px;" />
+<pre>
+<strong>Input:</strong> head = [1,2]
+<strong>Output:</strong> [2,1]
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> head = []
+<strong>Output:</strong> []
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the list is the range <code>[0, 5000]</code>.</li>
+	<li><code>-5000 &lt;= Node.val &lt;= 5000</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong> A linked list can be reversed either iteratively or recursively. Could you implement both?</p>

Solutions/0206-reverse-linked-list/solution.py

@@ -0,0 +1,15 @@
+# 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]:
+        dummy = ListNode()
+        current = head
+        while current:
+            next_node = current.next
+            current.next = dummy.next
+            dummy.next = current
+            current = next_node
+        return dummy.next