Merge two unsorted linked lists to get a sorted list - GeeksforGeeks

27 Dec, 2021
Given two unsorted Linked List, the task is to merge them to get a sorted singly linked list.Examples: 
Input: List 1 = 3 -> 1 -> 5, List 2 = 6-> 2 -> 4 Output: 1 -> 2 -> 3 -> 4 -> 5 -> 6
Input: List 1 = 4 -> 7 -> 5, List 2 = 2-> 1 -> 8 -> 1 Output: 1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8  
Naive Approach: The naive approach is to sort the given linked lists and then merge the two sorted linked lists together into one list in increasing order.To solve the problem mentioned above the naive method is to sort the two linked lists individually and merge the two linked lists together into one list which is in increasing order.
Efficient Approach: To optimize the above method we will concatenate the two linked lists and then sort it using any sorting algorithm. Below are the steps: 
Concatenate the two lists by traversing the first list until we reach it’s a tail node and then point the next of the tail node to the head node of the second list. Store this concatenated list in the first list.Sort the above-merged linked list. Here, we will use a bubble sort. So, if node->next->data is less then node->data, then swap the data of the two adjacent nodes.
Concatenate the two lists by traversing the first list until we reach it’s a tail node and then point the next of the tail node to the head node of the second list. Store this concatenated list in the first list.
Sort the above-merged linked list. Here, we will use a bubble sort. So, if node->next->data is less then node->data, then swap the data of the two adjacent nodes.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Create structure for a nodestruct node {    int data;    node* next;}; // Function to print the linked listvoid setData(node* head){    node* tmp;     // Store the head of the linked    // list into a temporary node*    // and iterate    tmp = head;     while (tmp != NULL) {         cout << tmp->data             << " -> ";        tmp = tmp->next;    }} // Function takes the head of the// LinkedList and the data as// argument and if no LinkedList// exists, it creates one with the// head pointing to first node.// If it exists already, it appends// given node at end of the last nodenode* getData(node* head, int num){     // Create a new node    node* temp = new node;    node* tail = head;     // Insert data into the temporary    // node and point it's next to NULL    temp->data = num;    temp->next = NULL;     // Check if head is null, create a    // linked list with temp as head    // and tail of the list    if (head == NULL) {        head = temp;        tail = temp;    }     // Else insert the temporary node    // after the tail of the existing    // node and make the temporary node    // as the tail of the linked list    else {         while (tail != NULL) {             if (tail->next == NULL) {                tail->next = temp;                tail = tail->next;            }            tail = tail->next;        }    }     // Return the list    return head;} // Function to concatenate the two listsnode* mergelists(node** head1,                 node** head2){     node* tail = *head1;     // Iterate through the head1 to find the    // last node join the next of last node    // of head1 to the 1st node of head2    while (tail != NULL) {         if (tail->next == NULL            && head2 != NULL) {            tail->next = *head2;            break;        }        tail = tail->next;    }     // return the concatenated lists as a    // single list - head1    return *head1;} // Sort the linked list using bubble sortvoid sortlist(node** head1){    node* curr = *head1;    node* temp = *head1;     // Compares two adjacent elements    // and swaps if the first element    // is greater than the other one.    while (curr->next != NULL) {         temp = curr->next;        while (temp != NULL) {             if (temp->data < curr->data) {                int t = temp->data;                temp->data = curr->data;                curr->data = t;            }            temp = temp->next;        }        curr = curr->next;    }} // Driver Codeint main(){    node* head1 = new node;    node* head2 = new node;     head1 = NULL;    head2 = NULL;     // Given Linked List 1    head1 = getData(head1, 4);    head1 = getData(head1, 7);    head1 = getData(head1, 5);     // Given Linked List 2    head2 = getData(head2, 2);    head2 = getData(head2, 1);    head2 = getData(head2, 8);    head2 = getData(head2, 1);     // Merge the two lists    // in a single list    head1 = mergelists(&head1,                       &head2);     // Sort the unsorted merged list    sortlist(&head1);     // Print the final    // sorted merged list    setData(head1);    return 0;}
// Java program for// the above approachclass GFG{ static node head1 = null;static node head2 = null;   // Create structure for a nodestatic class node{  int data;  node next;}; // Function to print// the linked liststatic void setData(node head){  node tmp;     // Store the head of the linked  // list into a temporary node  // and iterate  tmp = head;   while (tmp != null)  {    System.out.print(tmp.data + " -> ");    tmp = tmp.next;  }} // Function takes the head of the// LinkedList and the data as// argument and if no LinkedList// exists, it creates one with the// head pointing to first node.// If it exists already, it appends// given node at end of the last nodestatic node getData(node head, int num){  // Create a new node  node temp = new node();  node tail = head;   // Insert data into the temporary  // node and point it's next to null  temp.data = num;  temp.next = null;   // Check if head is null, create a  // linked list with temp as head  // and tail of the list  if (head == null)  {    head = temp;    tail = temp;  }   // Else insert the temporary node  // after the tail of the existing  // node and make the temporary node  // as the tail of the linked list  else  {    while (tail != null)    {      if (tail.next == null)      {        tail.next = temp;        tail = tail.next;      }      tail = tail.next;    }  }   // Return the list  return head;} // Function to concatenate// the two listsstatic node mergelists(){  node tail = head1;   // Iterate through the  // head1 to find the  // last node join the  // next of last node  // of head1 to the  // 1st node of head2  while (tail != null)  {    if (tail.next == null &&        head2 != null)    {      tail.next = head2;      break;    }    tail = tail.next;  }   // return the concatenated  // lists as a single list - head1  return head1;} // Sort the linked list// using bubble sortstatic void sortlist(){  node curr = head1;  node temp = head1;   // Compares two adjacent elements  // and swaps if the first element  // is greater than the other one.  while (curr.next != null)  {    temp = curr.next;    while (temp != null)    {      if (temp.data < curr.data)      {        int t = temp.data;        temp.data = curr.data;        curr.data = t;      }      temp = temp.next;    }    curr = curr.next;  }} // Driver Codepublic static void main(String[] args){  // Given Linked List 1  head1 = getData(head1, 4);  head1 = getData(head1, 7);  head1 = getData(head1, 5);   // Given Linked List 2  head2 = getData(head2, 2);  head2 = getData(head2, 1);  head2 = getData(head2, 8);  head2 = getData(head2, 1);   // Merge the two lists  // in a single list  head1 = mergelists();   // Sort the unsorted merged list  sortlist();   // Print the final  // sorted merged list  setData(head1);}} // This code is contributed by shikhasingrajput
# Python3 program for the# above approach # Create structure for a nodeclass node:       def __init__(self, x):               self.data = x        self.next = None # Function to print the linked# listdef setData(head):     # Store the head of the    # linked list into a    # temporary node* and    # iterate    tmp = head     while (tmp != None):        print(tmp.data,              end = " -> ")        tmp = tmp.next # Function takes the head of the# LinkedList and the data as# argument and if no LinkedList# exists, it creates one with the# head pointing to first node.# If it exists already, it appends# given node at end of the last nodedef getData(head, num):     # Create a new node    temp = node(-1)    tail = head     # Insert data into the temporary    # node and point it's next to NULL    temp.data = num    temp.next = None     # Check if head is null, create a    # linked list with temp as head    # and tail of the list    if (head == None):        head = temp        tail = temp     # Else insert the temporary node    # after the tail of the existing    # node and make the temporary node    # as the tail of the linked list    else:        while (tail != None):            if (tail.next == None):                tail.next = temp                tail = tail.next            tail = tail.next     # Return the list    return head # Function to concatenate the# two listsdef mergelists(head1,head2):     tail = head1     # Iterate through the head1 to    # find the last node join the    # next of last node of head1    # to the 1st node of head2    while (tail != None):        if (tail.next == None            and head2 != None):            tail.next =head2            break        tail = tail.next     # return the concatenated    # lists as a single list    # - head1    return head1 # Sort the linked list using# bubble sortdef sortlist(head1):       curr = head1    temp = head1     # Compares two adjacent elements    # and swaps if the first element    # is greater than the other one.    while (curr.next != None):        temp = curr.next        while (temp != None):            if (temp.data < curr.data):                t = temp.data                temp.data = curr.data                curr.data = t            temp = temp.next        curr = curr.next # Driver Codeif __name__ == '__main__':       head1 = node(-1)    head2 = node(-1)     head1 = None    head2 = None     # Given Linked List 1    head1 = getData(head1, 4)    head1 = getData(head1, 7)    head1 = getData(head1, 5)     # Given Linked List 2    head2 = getData(head2, 2)    head2 = getData(head2, 1)    head2 = getData(head2, 8)    head2 = getData(head2, 1)     # Merge the two lists    # in a single list    head1 = mergelists(head1,head2)     # Sort the unsorted merged list    sortlist(head1)     # Print the final    # sorted merged list    setData(head1) # This code is contributed by Mohit Kumar 29
// C# program for// the above approachusing System; class GFG{ static node head1 = null;static node head2 = null;   // Create structure for a nodeclass node{    public int data;    public node next;}; // Function to print// the linked liststatic void setData(node head){    node tmp;         // Store the head of the linked    // list into a temporary node    // and iterate    tmp = head;         while (tmp != null)    {        Console.Write(tmp.data + " -> ");        tmp = tmp.next;    }} // Function takes the head of//the List and the data as// argument and if no List// exists, it creates one with the// head pointing to first node.// If it exists already, it appends// given node at end of the last nodestatic node getData(node head, int num){         // Create a new node    node temp = new node();    node tail = head;         // Insert data into the temporary    // node and point it's next to null    temp.data = num;    temp.next = null;         // Check if head is null, create a    // linked list with temp as head    // and tail of the list    if (head == null)    {        head = temp;        tail = temp;    }         // Else insert the temporary node    // after the tail of the existing    // node and make the temporary node    // as the tail of the linked list    else    {        while (tail != null)        {            if (tail.next == null)            {                tail.next = temp;                tail = tail.next;            }            tail = tail.next;        }    }         // Return the list    return head;} // Function to concatenate// the two listsstatic node mergelists(){    node tail = head1;         // Iterate through the    // head1 to find the    // last node join the    // next of last node    // of head1 to the    // 1st node of head2    while (tail != null)    {        if (tail.next == null &&                head2 != null)        {            tail.next = head2;            break;        }        tail = tail.next;    }         // return the concatenated    // lists as a single list - head1    return head1;} // Sort the linked list// using bubble sortstatic void sortlist(){    node curr = head1;    node temp = head1;         // Compares two adjacent elements    // and swaps if the first element    // is greater than the other one.    while (curr.next != null)    {        temp = curr.next;        while (temp != null)        {            if (temp.data < curr.data)            {                int t = temp.data;                temp.data = curr.data;                curr.data = t;            }            temp = temp.next;        }        curr = curr.next;    }} // Driver Codepublic static void Main(String[] args){         // Given Linked List 1    head1 = getData(head1, 4);    head1 = getData(head1, 7);    head1 = getData(head1, 5);         // Given Linked List 2    head2 = getData(head2, 2);    head2 = getData(head2, 1);    head2 = getData(head2, 8);    head2 = getData(head2, 1);         // Merge the two lists    // in a single list    head1 = mergelists();         // Sort the unsorted merged list    sortlist();         // Print the final    // sorted merged list    setData(head1);}} // This code is contributed by Amit Katiyar
<script>// javascript program for// the above approach       var head1 = null;    var head2 = null;     // Create structure for a node    class node {    constructor(){         this.data=0;         this.next = null;        }    }     // Function to print    // the linked list    function setData( head) {        var tmp;         // Store the head of the linked        // list into a temporary node        // and iterate        tmp = head;         while (tmp != null) {            document.write(tmp.data + " -> ");            tmp = tmp.next;        }    }     // Function takes the head of the    // LinkedList and the data as    // argument and if no LinkedList    // exists, it creates one with the    // head pointing to first node.    // If it exists already, it appends    // given node at end of the last node    function getData( head , num) {        // Create a new node         temp = new node();        var tail = head;         // Insert data into the temporary        // node and point it's next to null        temp.data = num;        temp.next = null;         // Check if head is null, create a        // linked list with temp as head        // and tail of the list        if (head == null) {            head = temp;            tail = temp;        }         // Else insert the temporary node        // after the tail of the existing        // node and make the temporary node        // as the tail of the linked list        else {            while (tail != null) {                if (tail.next == null) {                    tail.next = temp;                    tail = tail.next;                }                tail = tail.next;            }        }         // Return the list        return head;    }     // Function to concatenate    // the two lists    function mergelists() {         tail = head1;         // Iterate through the        // head1 to find the        // last node join the        // next of last node        // of head1 to the        // 1st node of head2        while (tail != null) {            if (tail.next == null && head2 != null) {                tail.next = head2;                break;            }            tail = tail.next;        }         // return the concatenated        // lists as a single list - head1        return head1;    }     // Sort the linked list    // using bubble sort    function sortlist() {         curr = head1;         temp = head1;         // Compares two adjacent elements        // and swaps if the first element        // is greater than the other one.        while (curr.next != null) {            temp = curr.next;            while (temp != null) {                if (temp.data < curr.data) {                    var t = temp.data;                    temp.data = curr.data;                    curr.data = t;                }                temp = temp.next;            }            curr = curr.next;        }    }     // Driver Code             // Given Linked List 1        head1 = getData(head1, 4);        head1 = getData(head1, 7);        head1 = getData(head1, 5);         // Given Linked List 2        head2 = getData(head2, 2);        head2 = getData(head2, 1);        head2 = getData(head2, 8);        head2 = getData(head2, 1);         // Merge the two lists        // in a single list        head1 = mergelists();         // Sort the unsorted merged list        sortlist();         // Print the final        // sorted merged list        setData(head1); // This code is contributed by umadevi9616.</script>
1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8
 
Time Complexity: O((M+N)^2) where M and N are the lengths of the two given linked lists. We are merging the two list and performing bubble sort on the merged list. The time complexity of bubble sort is quadratic.Auxiliary Space: O(1)
shikhasingrajput
amit143katiyar
mohit kumar 29
umadevi9616
pankajsharmagfg
abhinavjain194
simranarora5sos
simmytarika5
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