Create 21.MergeTwoSortedLists.py

Author: YuriSpiridonov

Easy/21.MergeTwoSortedLists.py

@@ -0,0 +1,55 @@
+'''
+    Merge two sorted linked lists and return it as a new sorted 
+    list. The new list should be made by splicing together the 
+    nodes of the first two lists.
+
+    Example:
+    Input: l1 = [1,2,4], l2 = [1,3,4]
+    Output: [1,1,2,3,4,4]
+
+    Example:
+    Input: l1 = [], l2 = []
+    Output: []
+
+    Example:
+    Input: l1 = [], l2 = [0]
+    Output: [0]
+
+    Constraints:
+        - The number of nodes in both lists is in the range 
+          [0, 50].
+        - -100 <= Node.val <= 100
+        - Both l1 and l2 are sorted in non-decreasing order.
+'''
+#Difficulty: Easy
+#208 / 208 test cases passed.
+#Runtime: 40 ms
+#Memory Usage: 14.3 MB
+
+#Runtime: 40 ms, faster than 40.63% of Python3 online submissions for Merge Two Sorted Lists.
+#Memory Usage: 14.3 MB, less than 26.66% of Python3 online submissions for Merge Two Sorted Lists.
+
+# 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, l1: ListNode, l2: ListNode) -> ListNode:
+        node1 = l1
+        node2 = l2
+        l3 = ListNode()
+        node = l3
+        while node:
+            if node1 and node2 and node1.val > node2.val:
+                node.next = node2
+                node2 = node2.next
+            elif node1:
+                node.next = node1
+                node1 = node1.next
+            elif node2:
+                node.next = node2
+                node2 = node2.next
+            node = node.next
+        return l3.next