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 :



title-img


                            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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