From 14d5cc2e853f46cf4aa1d4c4a5e9d193d32a824a Mon Sep 17 00:00:00 2001
From: Selena Xiao <55109903+selenaxiao@users.noreply.github.com>
Date: Mon, 9 May 2022 17:58:21 -0400
Subject: [PATCH] python3 solution to #21
---
python3/21.py | 30 ++++++++++++++++++++++++++++++
1 file changed, 30 insertions(+)
create mode 100644 python3/21.py
diff --git a/python3/21.py b/python3/21.py
new file mode 100644
index 0000000000..62cd3a807a
--- /dev/null
+++ b/python3/21.py
@@ -0,0 +1,30 @@
+
+# https://leetcode.com/problems/merge-two-sorted-lists/
+# Iterative solution
+class Solution:
+ def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
+ tempHead = ListNode() # Empty node to be head of merged list
+ temp = tempHead # End node of merged list, append new nodes to end of temp
+
+ # While BOTH l1 and l2 still have nodes
+ while (l1 is not None and l2 is not None):
+ # If current node of l1 has a smaller value than current node of l2
+ if (l1.val < l2.val):
+ # Append l1 node to merged list
+ temp.next = l1
+ # Go to next l1 node
+ l1 = l1.next
+ else:
+ temp.next = l2
+ l2 = l2.next
+
+ # Go to new end of merged list
+ temp = temp.next
+
+ # Either l1, l2, or neither have nodes left
+ # Can directly append all of the remaining nodes to merged list
+ temp.next = l1 if l1 is not None else l2
+
+ # tempHead is empty node, return the actual head node of merged list
+ return tempHead.next
+
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