### Problem Statement#

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.

#### Example 1:#

``````"""
1-->2-->4-->5
^list1

1-->3-->4
^list2

1-->1-->2-->3-->4-->4-->5

^curr
"""
Input: list1 = [1,2,4,5], list2 = [1,3,4]
Output: [1,1,2,3,4,4,5]
``````

#### Example 2:#

``````Input: list1 = [], list2 = []
Output: []
``````

#### Example 3:#

``````Input: list1 = [], list2 = 
Output: 
``````

### Solution#

#### Intuition & Patterns#

In order to keep track of the head of the merged list, we will create a `dummy` node which will function as a `pre-head` node. Once we have completed the merge operation, we will return `dummy.next` to return the actual head of the merged linked lists.

Since we need to save the memory address of the `dummy` variable, we will create a `curr` node that we can use to modify the next pointers along the path. While `list1` and `list2` nodes exist, we traverse through the `linked lists` and set the `next` pointer to the next smallest node. Once one of the nodes becomes `None`, we set `curr.next` to the remaining list. Returning `dummy.next` gives us the head of the merged list. #### Python Code#

``````# 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]:
"""
T: O(N + M)
SL O(1)
"""
dummy = ListNode(-1)
curr = dummy

while list1 and list2:
if list1.val < list2.val:
curr.next = list1
list1 = list1.next
else:
curr.next = list2
list2 = list2.next
curr = curr.next

if list1:
curr.next = list1
else:
curr.next = list2

return dummy.next
``````

Leetcode 21: Merge Two Sorted Lists