diff --git a/work/1/Easy/Linked List/21.merge-two-sorted-lists.py b/work/1/Easy/Linked List/21.merge-two-sorted-lists.py
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+"""
+21. Merge Two Sorted Lists
+Difficulty: Easy
+https://leetcode.com/problems/merge-two-sorted-lists/
+
+──────────────────────────────────────────────────
+
+You are given the heads of two sorted linked lists list1 and list2.
+
+Merge the two lists into one sorted list. The list should be made by
+splicing together the nodes of the first two lists.
+
+Return the head of the merged linked list.
+
+
+
+Example 1:
+
+Input: list1 = [1,2,4], list2 = [1,3,4]
+Output: [1,1,2,3,4,4]
+
+Example 2:
+
+Input: list1 = [], list2 = []
+Output: []
+
+Example 3:
+
+Input: list1 = [], list2 = [0]
+Output: [0]
+
+
+
+Constraints:
+
+        • The number of nodes in both lists is in the range [0, 50].
+
+        • -100 <= Node.val <= 100
+
+        • Both list1 and list2 are sorted in non-decreasing order.
+"""
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(
+        self, list1: Optional[ListNode], list2: Optional[ListNode]
+    ) -> Optional[ListNode]:
+
+        if not list1:
+            return list2
+        if not list2:
+            return list1
+
+        if list1.val < list2.val:
+            head = list1
+            list1 = list1.next
+        else:
+            head = list2
+            list2 = list2.next
+
+        current = head
+        while list1 and list2:
+            if list1.val < list2.val:
+                current.next = list1
+                list1 = list1.next
+            else:
+                current.next = list2
+                list2 = list2.next
+            current = current.next
+
+        current.next = list1 or list2
+        return head