### Introduction

The linked list is one of the most important concepts to know while preparing for interviews. Having a good grasp of a linked list can be a huge plus point in coding interviews.

### Problem Statement

In this problem, we will be given two sorted linked lists. We need to merge both lists such that the newly created list is also in sorted order.

### Problem Statement Understanding

The problem statement is quite straightforward, we will be given two linked lists that are sorted in nature, and then we need to form a linked list using all the nodes of both linked lists such that the newly formed list is sorted in order.

Let’s try to understand the problem with the help of examples.

- Let the two sorted lists given to us be:

- Now, the list that we need to return must contain all the nodes of both lists in sorted order.
- So, the newly formed resultant list will be:

Letâ€™s take another example: - Let the two sorted lists given to us be 2â†’8â†’9â†’10â†’NULL and 11â†’13â†’17â†’NULL
- For the above two sorted lists, the final linked list after merging will be 2â†’8â†’9â†’10â†’11â†’13â†’17â†’NULL

##### Some more examples

Sample Input 1:

**list 1:**1â†’3â†’5â†’7**list 2:**2â†’4â†’6â†’8

Sample Output 1:

1â†’2â†’3â†’4â†’5â†’6â†’7â†’8

Sample Input 2:

**list 1:**2â†’3â†’6â†’7â†’9**list 2:**1â†’5â†’8â†’11â†’19â†’31

Sample Output 2:

1â†’2â†’3â†’5â†’6â†’7â†’8â†’9â†’11â†’19â†’31

Now, I think from the above examples, the problem statement is clear. Letâ€™s see how we can approach it.

Before moving to the approach section, try to think how you can approach this problem.

- If stuck, no problem, we will thoroughly see how we can approach this problem in the next section.

Letâ€™s move to the approach section.

### Approach 1

- We will make a recursive function that will create a sorted list out of two given sorted lists, containing all the nodes of both the given sorted list.
- We will compare the head of both the lists and the one with a smaller head will be appearing at the current position and all other nodes will appear after that.
- Now we will call the recursive function for the rest of the remaining nodes, and then we will attach the current node with the result from the recursive call, i.e., with the remaining nodeâ€™s result.
- If any one of the nodes is NULL, then we need to return the other listâ€™s head from there.

**Time Complexity:** O(n), n is the number of nodes in both the list combined.

**Space Complexity:** O(n), n is the size of the recursive stack

The above approach works fine, but can we make our code more efficient in terms of memory.

- The answer is yes, and to know how we can reduce the space complexity, read the below approach which uses constant memory.

Letâ€™s jump to the next approach.

### Approach 2

- In this approach, we first reverse both the lists given to us as input.
- Then we will create an empty list
**result**. - We will initialize two pointers
**h1**and**h2**to the head of**list 1**and**list 2**, respectively. - Then we will iterate over both the lists
**while(h1 != NULL && h2!=NULL)**.- While iterating at every step, we will find the larger of the two
**h1**and**h2**by comparing**h1â†’data**and**h2â†’data**. - We will insert the larger one of
**h1**and**h2**at the front of the result list. - Then we will move ahead in the list whose node was larger (larger node – we will compare using
**h1â†’data**and**h2â†’data**).

- While iterating at every step, we will find the larger of the two
- If any of the
**h1**or**h2**becomes NULL, we will insert all the remaining nodeâ€™s of the other list into the**result**list. - At last, we will have a new list that will be sorted given by
**result**.

**Time Complexity:** O(n), n is the number of nodes in both the list combined.

**Space Complexity:** O(1)

The above approach uses constant space, but it will iterate on both lists two times in the worst case.

- First, when the list is reversed and the second time when we are iterating to sort the list.

Can we do even better?

- The answer is yes, and to know the solution, read the below approach.

### Approach 3

- In this approach, we create a
**dummy node**with INT_MIN value that will represent the**tail**of our new list. - Then we will iterate on both the lists while both of them is not NULL.
- Then we need to check the node with a smaller value and insert it after the
**tail**of our new list, and then update the tail of the list. - At last, If the first list has reached NULL, then we need to attach the remaining elements of the second list after the
**tail**and vice versa.

### Algorithm

- Create a node named
**dummy**having INT_MIN as its data. - Initialize a pointer named
**tail**with the address of the above-created**dummy**variable. - Initialize two pointers named
**l1**and**l2**with the heads of both the lists, respectively. - Until both
**l1**and**l2**are not equal to NULL, run a while loop in which we need to check which pointer points to a smaller node.- If
**l1**points to a smaller node, we need to attach**l1**at the end of the tail and update**l1**with its next nodeâ€™s address. - If
**l2**points to a smaller node, we need to attach**l2**at the end of the tail and update**l2**with its next nodeâ€™s address - Then, we need to advance the
**tail**pointer to its next node

- If
- After the while loop ends, we need to check whether
**l1**or**l2**is NULL or not.- If
**l2**is NULL, we need to assign**tailâ†’next**to**l1**. - If
**l1**is NULL, we need to assign**tailâ†’next**to**l1**.

- If
- Finally, we return the
**dummy nodeâ€™s next node**as the head of our newly created list.

### Dry Run

### Code Implementation

#include<stdio.h> #include<stdlib.h> #include<assert.h> /* Link list node */ struct Node { int data; struct Node* next; }; /* pull off the front node of the source and put it in dest */ void MoveNode(struct Node** destRef, struct Node** sourceRef); /* Takes two lists sorted in increasing order, and splices their nodes together to make one big sorted list which is returned. */ struct Node* SortedMerge(struct Node* a, struct Node* b) { /* a dummy first node to hang the result on */ struct Node dummy; /* tail points to the last result node */ struct Node* tail = &dummy; /* so tail->next is the place to add new nodes to the result. */ dummy.next = NULL; while (1) { if (a == NULL) { /* if either list runs out, use the other list */ tail->next = b; break; } else if (b == NULL) { tail->next = a; break; } if (a->data <= b->data) MoveNode(&(tail->next), &a); else MoveNode(&(tail->next), &b); tail = tail->next; } return(dummy.next); } /* UTILITY FUNCTIONS */ /* MoveNode() function takes the node from the front of the source, and move it to the front of the dest. It is an error to call this with the source list empty. Before calling MoveNode(): source == {1, 2, 3} dest == {1, 2, 3} After calling MoveNode(): source == {2, 3} dest == {1, 1, 2, 3} */ void MoveNode(struct Node** destRef, struct Node** sourceRef) { /* the front source node */ struct Node* newNode = *sourceRef; assert(newNode != NULL); /* Advance the source pointer */ *sourceRef = newNode->next; /* Link the old dest off the new node */ newNode->next = *destRef; /* Move dest to point to the new node */ *destRef = newNode; } /* Function to insert a node at the beginning of the linked list */ void push(struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList(struct Node *node) { while (node!=NULL) { printf("%d ", node->data); node = node->next; } } /* Driver program to test above functions*/ int main() { /* Start with the empty list */ struct Node* res = NULL; struct Node* a = NULL; struct Node* b = NULL; /* Let us create two sorted linked lists to test the functions Created lists, a: 5->10->15, b: 2->3->20 */ push(&a, 15); push(&a, 10); push(&a, 5); push(&b, 20); push(&b, 3); push(&b, 2); /* Remove duplicates from linked list */ res = SortedMerge(a, b); printf("Merged Linked List is: \n"); printList(res); return 0; }

#include <iostream> #include<bits/stdc++.h> using namespace std; /* Class with constructor for a node of Linked list */ class Node { public: int data; Node* next; // constructor Node(int x){ data = x; next = NULL; } }; /* Using this function we will be printing the data of the linked list */ void print(Node* head) { Node* curr = head; while (curr != NULL) { cout << curr->data << " -> "; curr = curr->next; } cout<<"NULL"<<endl; } /* Using this function we will be merging the two given sorted linked lists into a single sorted linked list */ Node *mergeTwoLists(Node *l1, Node *l2) { Node dummy(INT_MIN); Node *tail = &dummy; while (l1 && l2) { if (l1->data < l2->data) { tail->next = l1; l1 = l1->next; } else { tail->next = l2; l2 = l2->next; } tail = tail->next; } tail->next = l1 ? l1 : l2; return dummy.next; } /* Main function */ int main() { Node *list1 = new Node(2); list1->next = new Node(3); list1->next->next = new Node(5); list1->next->next->next = new Node(9); Node *list2 = new Node(1); list2->next = new Node(4); list2->next->next = new Node(8); cout<<"Original linked list 1: "<<endl; print(list1); cout<<"Original linked list 2: "<<endl; print(list2); Node* result = mergeTwoLists(list1,list2); cout<<"Merged sorted linked list: "<<endl; print(result); return 0; }

class Node { int data; Node next; Node(int d) {data = d; next = null;} } class Sort { Node sortedMerge(Node headA, Node headB) { /* a dummy first node to chang the result on */ Node dummyNode = new Node(0); /* tail points to the last result node */ Node tail = dummyNode; while(true) { /* if either list runs out, use the other list */ if(headA == null) { tail.next = headB; break; } if(headB == null) { tail.next = headA; break; } /* Compare the data of the two lists whichever lists' data is smaller, append it into tail and advance the head to the next Node*/ if(headA.data <= headB.data) { tail.next = headA; headA = headA.next; } else { tail.next = headB; headB = headB.next; } tail = tail.next; } return dummyNode.next; } } class Merge2Sorted { Node head; public void addToTheLast(Node node) { if (head == null) { head = node; } else { Node temp = head; while (temp.next != null) temp = temp.next; temp.next = node; } } void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } System.out.println(); } // Driver Code public static void main(String args[]) { Merge2Sorted llist1 = new Merge2Sorted(); Merge2Sorted llist2 = new Merge2Sorted(); // Node head1 = new Node(5); llist1.addToTheLast(new Node(5)); llist1.addToTheLast(new Node(10)); llist1.addToTheLast(new Node(15)); // Node head2 = new Node(2); llist2.addToTheLast(new Node(2)); llist2.addToTheLast(new Node(3)); llist2.addToTheLast(new Node(20)); llist1.head = new Sort().sortedMerge(llist1.head, llist2.head); llist1.printList(); } }

class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def printList(self): temp = self.head while temp: print(temp.data, end=" ") temp = temp.next def addToList(self, newData): newNode = Node(newData) if self.head is None: self.head = newNode return last = self.head while last.next: last = last.next last.next = newNode def mergeLists(headA, headB): dummyNode = Node(0) tail = dummyNode while True: if headA is None: tail.next = headB break if headB is None: tail.next = headA break if headA.data <= headB.data: tail.next = headA headA = headA.next else: tail.next = headB headB = headB.next tail = tail.next return dummyNode.next listA = LinkedList() listB = LinkedList() listA.addToList(2) listA.addToList(3) listA.addToList(5) listA.addToList(9) listB.addToList(1) listB.addToList(4) listB.addToList(8) print("Original linked list 1:") listA.printList() print("\nOriginal linked list 2:") listB.printList() listA.head = mergeLists(listA.head, listB.head) print("\nMerged Linked List is:") listA.printList()

#### Output

Original linked list 1:

2 -> 3 -> 5 -> 9 -> NULL

Original linked list 2:

1 -> 4 -> 8 -> NULL

Merged sorted linked list:

1 -> 2 -> 3 -> 4 -> 5 -> 8 -> 9 -> NULL

**Time Complexity:** O(n), where n is the total number of nodes in both the lists.

[forminator_quiz id=”5081″]

So, in this blog, we have tried to explain how you can merge two sorted linked lists in the most optimal way. This is a basic problem and is good for strengthening your concepts in LinkedList and if you want to practice more such problems, you can checkout Prepbytes (Linked List).