**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
```

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