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 -> 6Input: 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.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Create structure for a node struct node { int data; node* next; }; // Function to print the linked list void 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 node 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 lists node* 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 sort void 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 Code int 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
// Java program for // the above approach class GFG{ static node head1 = null ; static node head2 = null ; // Create structure for a node static class node { int data; node next; }; // Function to print // the linked list static 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 node static 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 lists static 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 sort static 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 Code public 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
# Python3 program for the # above approach # Create structure for a node class node: def __init__( self , x): self .data = x self . next = None # Function to print the linked # list def 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 node def 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 lists def 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 sort def 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 Code if __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#
// C# program for // the above approach using System; class GFG{ static node head1 = null ; static node head2 = null ; // Create structure for a node class node { public int data; public node next; }; // Function to print // the linked list static 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 node static 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 lists static 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 sort static 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 Code public 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 |
1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8
Time Complexity: O(M*N) where M and N are the lengths of the two given linked lists.