Merge two sorted linked lists

Write a SortedMerge() function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. SortedMerge() should return the new list. The new list should be made by splicing
together the nodes of the first two lists.

For example if the first linked list a is 5->10->15 and the other linked list b is 2->3->20, then SortedMerge() should return a pointer to the head node of the merged list 2->3->5->10->15->20.

There are many cases to deal with: either ‘a’ or ‘b’ may be empty, during processing either ‘a’ or ‘b’ may run out first, and finally there’s the problem of starting the result list empty, and building it up while going through ‘a’ and ‘b’.



Method 1 (Using Dummy Nodes)
The strategy here uses a temporary dummy node as the start of the result list. The pointer Tail always points to the last node in the result list, so appending new nodes is easy.
The dummy node gives tail something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’, and adding it to tail. When
we are done, the result is in dummy.next.

C/C++

/* C/C++ program to merge two sorted linked lists */
#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}
  
   Affter 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 beginging 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;
    }
}
  
/* Drier 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;
}

Java

/* Java program to merge two
   sorted linked lists */
import java.util.*;
  
/* Link list node */
class Node 
{
    int data;
    Node next;
    Node(int d) {data = d;
                 next = null;}
}
      
class MergeLists 
{
Node head; 
  
/* Method to insert a node at 
   the end of the linked list */
public void addToTheLast(Node node) 
{
    if (head == null)
    {
        head = node;
    }
    else 
    {
        Node temp = head;
        while (temp.next != null)
            temp = temp.next;
        temp.next = node;
    }
}
  
/* Method to print linked list */
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[])
{
    /* Let us create two sorted linked
       lists to test the methods 
       Created lists:
           llist1: 5->10->15,
           llist2: 2->3->20
    */
    MergeLists llist1 = new MergeLists();
    MergeLists llist2 = new MergeLists();
      
    // 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 Gfg().sortedMerge(llist1.head, 
                                        llist2.head);
    llist1.printList();  
      
}
}
  
class Gfg
{
/* Takes two lists sorted in 
increasing order, and splices 
their nodes together to make 
one big sorted list which is 
returned. */
Node sortedMerge(Node headA, Node headB)
{
      
    /* a dummy first node to 
       hang 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;
        }
          
        /* Advance the tail */
        tail = tail.next;
    }
    return dummyNode.next;
}
}
  
// This code is contributed
// by Shubhaw Kumar

Output :

Merged Linked List is: 
2 3 5 10 15 20 


Method 2 (Using Local References)
This solution is structurally very similar to the above, but it avoids using a dummy node. Instead, it maintains a struct node** pointer, lastPtrRef, that always points to the last pointer of the result list. This solves the same case that the dummy node did — dealing with the result list when it is empty. If you are trying to build up a list at its tail, either the dummy node or the struct node** “reference” strategy can be used (see Section 1 for details).

struct Node* SortedMerge(struct Node* a, struct Node* b) 
{
  struct Node* result = NULL;
    
  /* point to the last result pointer */
  struct Node** lastPtrRef = &result; 
    
  while(1) 
  {
    if (a == NULL) 
    {
      *lastPtrRef = b;
       break;
    }
    else if (b==NULL) 
    {
       *lastPtrRef = a;
       break;
    }
    if(a->data <= b->data) 
    {
      MoveNode(lastPtrRef, &a);
    }
    else 
    {
      MoveNode(lastPtrRef, &b);
    }
    
    /* tricky: advance to point to the next ".next" field */
    lastPtrRef = &((*lastPtrRef)->next); 
  }
  return(result);
}



Method 3 (Using Recursion)
Merge is one of those nice recursive problems where the recursive solution code is much cleaner than the iterative code. You probably wouldn’t want to use the recursive version for production code however, because it will use stack space which is proportional to the length of the lists.

C/C++

struct Node* SortedMerge(struct Node* a, struct Node* b) 
{
  struct 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 = SortedMerge(a->next, b);
  }
  else 
  {
     result = b;
     result->next = SortedMerge(a, b->next);
  }
  return(result);
}

Python3

# Python3 program merge two sorted linked
# in third linked list using recursive.
  
# Node class
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
  
  
# Constructor to initialize the node object
class LinkedList:
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # Method to print linked list
    def printList(self):
        temp = self.head
          
        while temp :
            print(temp.data, end="->")
            temp = temp.next
  
    # Function to add of node at the end.
    def append(self, new_data):
        new_node = Node(new_data)
          
        if self.head is None:
            self.head = new_node
            return
        last = self.head
          
        while last.next:
            last = last.next
        last.next = new_node
  
  
# Function to merge two sorted linked list.
def mergeLists(head1, head2):
  
    # create a temp node NULL
    temp = None
  
    # List1 is empty then return List2
    if head1 is None:
        return head2
  
    # if List2 is empty then return List1
    if head2 is None:
        return head1
  
    # If List1's data is smaller or
    # equal to List2's data
    if head1.data <= head2.data:
  
        # assign temp to List1's data
        temp = head1
  
        # Again check List1's data is smaller or equal List2's 
        # data and call mergeLists function.
        temp.next = mergeLists(head1.next, head2)
          
    else:
        # If List2's data is greater than or equal List1's 
        # data assign temp to head2
        temp = head2
  
        # Again check List2's data is greater or equal List's
        # data and call mergeLists function.
        temp.next = mergeLists(head1, head2.next)
  
    # return the temp list.
    return temp
  
  
# Driver Function
if __name__ == '__main__':
  
    # Create linked list :
    # 10->20->30->40->50
    list1 = LinkedList()
    list1.append(10)
    list1.append(20)
    list1.append(30)
    list1.append(40)
    list1.append(50)
  
    # Create linked list 2 :
    # 5->15->18->35->60
    list2 = LinkedList()
    list2.append(5)
    list2.append(15)
    list2.append(18)
    list2.append(35)
    list2.append(60)
  
    # Create linked list 3
    list3 = LinkedList()
  
    # Merging linked list 1 and linked list 2
    # in linked list 3
    list3.head = mergeLists(list1.head, list2.head)
  
    print(" Merged Linked List is : ", end="")
    list3.printList()     
  
  
# This code is contributed by 'Shriaknt13'.         

Please refer below post for simpler implementations :
Merge two sorted lists (in-place)

Source: http://cslibrary.stanford.edu/105/LinkedListProblems.pdf

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.



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Improved By : Shubhaw Kumar