### Merging Two Sorted Linked Lists using Java

Merging two sorted linked lists is a classic computer science problem often asked during technical interviews. The challenge is to combine the two lists into a single, sorted linked list without using any extra space.

# How the Algorithm Works

Two Pointers: We'll have two pointers, one for each list.

Compare: At each step, compare the current nodes of both lists and pick the node with the smaller value.

Move Forward: Move the pointer of the list from which the node was chosen to the next node.

Resulting List: The resulting list is constructed by combining the nodes in the required order.

Let's implement the Java program for the same.

# Java Program for Merging Two-Sorted Linked Lists

``````class Node {
int data;
Node next;

Node(int data) {
this.data = data;
this.next = null;
}
}

public class SortedListMerger {

public static Node mergeSortedLists(Node l1, Node l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;

// Initialize mergedList with a dummy node.
Node dummy = new Node(-1);
Node current = dummy;

while (l1 != null && l2 != null) {
if (l1.data < l2.data) {
current.next = l1;
l1 = l1.next;
} else {
current.next = l2;
l2 = l2.next;
}
current = current.next;
}

// If l1 is not exhausted, add remaining nodes of l1.
if (l1 != null) {
current.next = l1;
}

// If l2 is not exhausted, add remaining nodes of l2.
if (l2 != null) {
current.next = l2;
}

return dummy.next;  // Return next of dummy to skip the dummy node.
}

public static void main(String[] args) {
Node l1 = new Node(1);
l1.next = new Node(3);
l1.next.next = new Node(5);

Node l2 = new Node(2);
l2.next = new Node(4);
l2.next.next = new Node(6);

Node mergedList = mergeSortedLists(l1, l2);

System.out.println("Merged List:");
while (mergedList != null) {
System.out.print(mergedList.data + " ");
mergedList = mergedList.next;
}
}
}
``````

## Output:

``````Merged List:
1 2 3 4 5 6``````

## Explanation:

We start by defining our linked list's basic structure using the Node class.

The mergeSortedLists function handles the merging. It starts by checking edge cases - if either of the lists is null.

The dummy node acts as a placeholder to start building our merged list. This is helpful, so we don't have to handle the special case of initializing the head of the merged list.

As we iterate over l1 and l2, we determine which node should come next in the merged list based on the current nodes' values in l1 and l2. The current pointer keeps track of the last node in our merged list.

After processing all elements from either l1 or l2, we might still have some nodes left in one of the original lists. We append these remaining nodes to our merged list.

In the main method, we test our merging function. The merged list contains elements from both l1 and l2 in sorted order.

This efficient solution ensures that each node is processed once, making it a linear time O(n) algorithm, where n is the total number of nodes in both linked lists combined.

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