Solving Problem: Combine Two Linked Lists

Data Structures – Linked List

Problem Statement:

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into 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.

Example 1:

1 -> 2 -> 4 -> NULL

1 -> 3 -> 4 -> NULL

Expected Output:

1 -> 1 -> 2 -> 3 -> 4 -> 4 -> NULL

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

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

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

Constraints:

The number of nodes in both lists is in the range [0, 50].
-100 <= Node.val <= 100
Both list1 and list2 are sorted in non-decreasing order.

Solution:

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def mergeTwoLists(list1, list2):
        arr = []
        while list1 is not None:
            arr.append(list1.val)
            list1 = list1.next
        while list2 is not None:
            arr.append(list2.val)
            list2 = list2.next
        arr.sort()
        temp = ListNode(-1)
        curr = temp
        for value in arr:
            curr.next = ListNode(value)
            curr = curr.next
        return temp.next
        
        
def printList(node):
   while node:
       print(node.val, end=" -> ")
       node = node.next
   print("NULL")
if __name__ == "__main__":
  
    head1 = ListNode(1)
    head1.next = ListNode(2)
    head1.next.next = ListNode(4)
 
    head2 = ListNode(1)
    head2.next = ListNode(3)
    head2.next.next = ListNode(4)
    result = Solution.mergeTwoLists(head1, head2)
    printList(result)