21. Merge Two Sorted Lists

July 2023 | Solving the merge two sorted linked lists Leetcode 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]
Input: list1 = [], list2 = []
Output: []
Input: list1 = [], list2 = [0]
Output: [0]

Definition for singly-linked list:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


The original nodes have to be used. The "output list" is initially empty, so we can create a dummy node that points to the output list initially.

The dummy node will be initialized to list1 first, and we can start comparing the subsequent values:

Input: list1 = [1,2,4], list2 = [1,3,4]
dummy = node 1
# compare 1 (from list2) and 2 (from list1)
# insert node 1 (from list2) to end

When there are no more values in a list, simply add the nodes to the end of the output list (the provided lists are already sorted).


class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        # Pointer to beginning of output 
        dummy = ListNode()
        tail = dummy

        # We can compare values when BOTH lists still have a node 
        while list1 and list2:
            # Insert smaller value to tail
            if list1.val < list2.val:
                tail.next = list1
                list1 = list1.next
                tail.next = list2
                list2 = list2.next
            tail = tail.next
        # Check if a list is not empty
        if list1:
            tail.next = list1
        elif list2:
            tail.next = list2

        # return.next because dummy is just a pointer to the output
        return dummy.next