Add 'Reverse Linked List'

Author: nickolashkraus

README.md

@@ -32,6 +32,8 @@ A collection of LeetCode solutions
 
 [Product of Array Except Self](./src/product_of_array_except_self.py)
 
+[Reverse Linked List](./src/reverse_linked_list.py)
+
 [Same Tree](./src/same_tree.py)
 
 [String Compression](./src/string_compression.py)

TODO.md

@@ -8,3 +8,4 @@
 - [ ] Add alternative solution for 'Longest Increasing Subsequence'
 - [ ] Add binary search solution for 'Longest Increasing Subsequence'
 - [ ] Ensure all lines are < 80 characters.
+- [ ] Add recursive solution to 'Reverse Linked List'

src/reverse_linked_list.py

@@ -0,0 +1,67 @@
+"""
+206. Reverse Linked List
+
+https://leetcode.com/problems/reverse-linked-list
+
+NOTES
+  * Reversing a linked list is similar to merging two sorted linked lists, in
+    that a pointer ('prev') keeps a reference to the previous node. However, we
+    also need an additional pointer ('next') to keep a reference to the next
+    node in the sequence.
+
+    Example:
+
+      1   2   3   4   5
+      ● → ● → ● → ● → ● → ○
+      ↑
+
+      Starting with head (1), we store its reference to next (2):
+
+        next = curr.next
+
+      Next, we update its reference to next to be the previous node (null):
+
+        curr.next = prev
+
+          1   2   3   4   5
+      ○ ← ●   ● → ● → ● → ● → ○
+
+      Then, we update the previous node (null) to be the current node (1):
+
+        prev = curr
+
+      Finally, we update the current node to be the next node (2):
+
+          1   2   3   4   5
+      ○ ← ●   ● → ● → ● → ● → ○
+              ↑
+
+      This process is repeated until the current node is null:
+
+          1   2   3   4   5
+      ○ ← ● ← ● ← ● ← ● ← ●  ○
+                             ↑
+"""
+
+from src.classes import ListNode
+
+
+class Solution:
+    """
+    The time complexity for this solution is O(n) and the space complexity is
+    O(1).
+    """
+
+    def reverseList(self, head: ListNode | None) -> ListNode | None:
+        prev = None
+        curr = head
+        # 1. Store the current node's reference to next.
+        # 2. Update the current node's next to be the previous node.
+        # 3. Update the previous node to be the current node.
+        # 4. Update the current node to be the next node (stored in 1).
+        while curr:
+            _next = curr.next
+            curr.next = prev
+            prev = curr
+            curr = _next
+        return prev

tests/test_reverse_linked_list.py

@@ -0,0 +1,31 @@
+"""
+206. Reverse Linked List
+
+https://leetcode.com/problems/reverse-linked-list
+"""
+
+from unittest import TestCase
+
+from src.reverse_linked_list import Solution
+
+from .utils import create_linked_list_from_list, create_list_from_linked_list
+
+
+class TestSolution(TestCase):
+    def test_1(self):
+        exp = [5, 4, 3, 2, 1]
+        l = create_linked_list_from_list([1, 2, 3, 4, 5])
+        ll = Solution().reverseList(l)
+        assert create_list_from_linked_list(ll) == exp
+
+    def test_2(self):
+        exp = [2, 1]
+        l = create_linked_list_from_list([1, 2])
+        ll = Solution().reverseList(l)
+        assert create_list_from_linked_list(ll) == exp
+
+    def test_3(self):
+        exp = []
+        l = create_linked_list_from_list([])
+        ll = Solution().reverseList(l)
+        assert create_list_from_linked_list(ll) == exp