This code takes in two linked lists (`list1` and `list2`)...

September 4, 2025 at 05:08 PM

public static void mergeTwoLists(Node list1, Node list2){
        Node head = null;
        Node n1 = list1;
        Node n2 = list2;
        if(n1.val<n2.val){
            head = new Node(n1.val,null);
            n1 = n1.next;
        }
        else{
            head = new Node(n2.val,null);
            n2 = n2.next;
        }
        Node temp = head;
        while(n1!=null && n2!=null){
            if(n1.val == n2.val){
                temp.next = n1;
                temp = temp.next;
                temp.next = n2;
                temp = temp.next;
                n1 = n1.next;
                n2 = n2.next;
            }
            else if(n1.val<n2.val){
                temp.next = n1;
                n1 = n1.next;
                temp = temp.next;
            }
            else{
                temp.next = n2;
                temp = temp.next;
                n2 = n2.next;
            }
        }

        while(n1!=null){
            temp.next = n1;
            n1 = n1.next;
            temp = temp.next;
        }

        while(n2!=null){
            temp.next = n2;
            temp = temp.next;
            n2 = n2.next;
        }
    }

This code takes in two linked lists (list1 and list2) and merges them into a single, sorted linked list in ascending order. The merged list includes the elements from both input lists while preserving their relative order.

Here's a more detailed explanation of what this code does:

Initialization:
- A reference to the merged linked list is created (head) and initially set to null.
- Two pointers (n1 and n2) are initialized to the heads of list1 and list2, respectively.
Setting the head of the merged list:
- The function compares the first nodes of the two lists (n1.val and n2.val).
- A new node is created with the smaller of the two values, and this node becomes the head of the merged list. The pointer (n1 or n2) to the smaller node is advanced to its next node.
Merging main part of the lists:
- The function iterates through n1 and n2 as long as both are not null.
- During each iteration:
  - If the values (n1.val and n2.val) are equal, both nodes are selected, one after the other, to preserve their order.
  - If n1.val is smaller, the node pointed to by n1 is added to the merged list, and n1 is advanced.
  - If n2.val is smaller, the node pointed to by n2 is added, and n2 is advanced.
Adding remaining nodes:
- If one of the lists (list1 or list2) is exhausted before the other, any remaining nodes from the non-exhausted list are appended to the merged list.
Result:
- The function merges the two input linked lists into a single sorted linked list but does not return anything (void return type). Instead, the merged result is constructed starting from the head variable.

Assumptions:

Each input linked list (list1 and list2) is already sorted in ascending order.
Node is a class with at least two fields: val (to hold the value) and next (to point to the next node in the list).

Example Scenario:

If list1 contains [1 -> 3 -> 5] and list2 contains [2 -> 4 -> 6], the merged linked list becomes [1 -> 2 -> 3 -> 4 -> 5 -> 6]. Similarly, if list1 contains [1 -> 3 -> 5] and list2 contains [3 -> 4 -> 6], the merged list would be [1 -> 3 -> 3 -> 4 -> 5 -> 6].

Note:

The head variable represents the start of the merged list, but since the merged list is constructed within the function and the function does not return the head, the merged list would not be accessible from outside unless the head or the merged list is explicitly passed or stored elsewhere.

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