submit 21 solution

Author: BrianRuizy

images/21 - Merge Two Sorted Lists.md

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+# 21 - Merge Two Sorted Lists
+
+Difficulty: easy
+Done: Yes
+Last edited: March 2, 2022 1:53 AM
+Link: https://leetcode.com/problems/merge-two-sorted-lists/submissions/
+Topic: linked list
+
+## Problem
+
+---
+
+You are given the heads of two sorted linked lists `list1` and `list2`. Merge the two lists in a 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*.
+
+```
+Input: list1 = [1,2,4], list2 = [1,3,4]
+Output: [1,1,2,3,4,4]
+```
+
+![Untitled](images/21-2.png)
+
+## Solution
+
+---
+
+Solution can be done in a single iterative traversal, resulting in linear time complexity. To do so we would need to traverse both lists with the use of a pointer —*previous—* which essentially tracks the current place in the final list. 
+
+Additionally we would need to add a sentinel node (hypothetical first node). This allows to functionally make the final merged list never empty.
+
+The while loop —typical for linked list problems— only runs while l1 or l2 do not point to None. Since 
+
+## Whiteboard
+
+---
+
+![Untitled](images/21-1.png)
+
+## Code
+
+---
+
+```python
+# 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]:
+        
+        # create sentinel node
+        s = ListNode(0, None)
+        previous = s
+        
+        while list1 is not None and list2 is not None:
+            # compare value at l1 vs value at l2
+            # whichever is smaller that's where we point previous, and step that list
+            # essentially prev keeps track of the new list
+            
+            if list1.val <= list2.val:
+                previous.next = list1
+                list1 = list1.next
+            else: 
+                previous.next = list2
+                list2 = list2.next
+                
+            previous = previous.next
+        
+        # the while loop only runs while l1 or l2 are not None
+        
+        # since lists can be diffent sizes,
+        # when we've reached the end for one we can point
+        # the merged list to the non-null list
+        previous.next = list1 or list2
+        
+        return s.next
+```
+
+## Time Complexity
+
+---
+
+big O