**Merge two sorted linked lists**

### Problem Statement :

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description Complete the mergeLists function in the editor below. mergeLists has the following parameters: SinglyLinkedListNode pointer headA: a reference to the head of a list SinglyLinkedListNode pointer headB: a reference to the head of a list Returns SinglyLinkedListNode pointer: a reference to the head of the merged list Input Format The first line contains an integer t, the number of test cases. The format for each test case is as follows: The first line contains an integer n, the length of the first linked list. The next n lines contain an integer each, the elements of the linked list. The next line contains an integer m , the length of the second linked list. The next m lines contain an integer each, the elements of the second linked list.

### Solution :

` ````
Solution in C :
In C++ :
/*
Merge two sorted lists A and B as one linked list
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
Node* MergeLists(Node *a, Node*b)
{
// This is a "method-only" submission.
// You only need to complete this method
Node* result = NULL;
/* Base cases */
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->next = MergeLists(a->next, b);
}
else
{
result = b;
result->next = MergeLists(a, b->next);
}
return(result);
}
In C :
// Complete the mergeLists function below.
/*
* For your reference:
*
* SinglyLinkedListNode {
* int data;
* SinglyLinkedListNode* next;
* };
*
*/
SinglyLinkedListNode* mergeLists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
SinglyLinkedList *newHead = malloc(sizeof(SinglyLinkedList));
newHead->head = NULL;
newHead->tail = NULL;
while(head1 != NULL && head2 != NULL){
if(head1->data > head2->data){
insert_node_into_singly_linked_list(&newHead, head2->data);
head2 = head2->next;
}
else if(head1->data < head2->data){
insert_node_into_singly_linked_list(&newHead, head1->data);
head1 = head1->next;
}
else{
insert_node_into_singly_linked_list(&newHead, head1->data);
insert_node_into_singly_linked_list(&newHead, head2->data);
head1 = head1->next;
head2 = head2->next;
}
}
while(head1 != NULL){
insert_node_into_singly_linked_list(&newHead, head1->data);
head1 = head1->next;
}
while(head2 != NULL){
insert_node_into_singly_linked_list(&newHead, head2->data);
head2 = head2->next;
}
return newHead->head;
}
In Java :
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
}
*/
Node MergeLists(Node list1, Node list2) {
// This is a "method-only" submission.
// You only need to complete this method
Node dummy = new Node();
dummy.next=null;
dummy.data=0;
Node temp= dummy;
while(true)
{
if(list1==null)
{ temp.next = list2;
break;
}
else if(list2==null){
temp.next = list1;
break;
}
else if(list1.data < list2.data)
{
temp.next=list1;
list1=list1.next;
}
else{
temp.next=list2;
list2=list2.next;
}
temp=temp.next;
}
return dummy.next;
}
In python3 :
"""
Merge two linked lists
head could be None as well for empty list
Node is defined as
class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
return back the head of the linked list in the below method.
"""
def MergeLists(headA, headB):
if (headA==None) | (headB==None):
if(headA == None):
return headB
else:
return headA
if headA.data > headB.data: # headA points to the smaller one
temp = headA
headA = headB
headB = temp
head = headA
curA = headA.next # curA one step ahead
curB = headB
while (curA!=None) & (curB!=None):
if (curA.data>=curB.data):
headB =headB.next
curB.next = curA
headA.next = curB
headA = curA
curA = curA.next
curB = headB
else:
headA =curA
curA = curA.next
if curA == None:
headA.next = headB
return head
```

## View More Similar Problems

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

View Solution →## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

View Solution →## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

View Solution →## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

View Solution →## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

View Solution →## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →