Merge two sorted linked lists
AuxiliaryGiven two sorted linked lists consisting of N and M nodes respectively. The task is to merge both of the lists (in place) and return the head of the merged list.
Examples:
Input: a: 5->10->15, b: 2->3->20
Output: 2->3->5->10->15->20Input: a: 1->1, b: 2->4
Output: 1->1->2->4
Brute Force Way:
The Approach:
In this Approach we use a vector of (n+m) size where n and m are length respective linked list and then store all the element in vector and then we sort the vector and make new linked list it will be our answer.
C++
#include <bits/stdc++.h>using namespace std; /* Link list Node */struct Node { int key; struct Node* next;};Node* newNode(int key){ Node* temp = new Node; temp->key = key; temp->next = NULL; return temp;}int main() { /* Let us create two sorted linked lists to test the above functions. Created lists shall be a: 5->10->15->40 b: 2->3->20 */ Node* a = newNode(5); a->next = newNode(10); a->next->next = newNode(15); a->next->next->next = newNode(40); Node* b = newNode(2); b->next = newNode(3); b->next->next = newNode(20); vector<int>v; while(a!=NULL){ v.push_back(a->key); a=a->next; } while(b!=NULL){ v.push_back(b->key); b=b->next; } sort(v.begin(),v.end()); Node* result= newNode(-1); Node* temp=result; for(int i=0;i<v.size();i++){ result->next=newNode(v[i]); result=result->next; } temp=temp->next; cout<<"Resultant Merge Linked List Is :"<<endl; while(temp!=NULL){ cout<<temp->key<<" "; temp=temp->next; } return 0;} |
Java
import java.util.ArrayList;import java.util.Collections;import java.util.List;// link list nodeclass Node { int key; Node next; public Node(int key) { this.key = key; next = null; }}public class Main { // return a newnode public static Node newNode(int key) { return new Node(key); } // driver code public static void main(String[] args) { // let us create two sorted linked lists to test the above // function. Created lists shall be // a: 5->10->15->40 // b: 2->3->20 Node a = new Node(5); a.next = new Node(10); a.next.next = new Node(15); a.next.next.next = new Node(40); Node b = new Node(2); b.next = new Node(3); b.next.next = new Node(20); List<Integer> v = new ArrayList<>(); while (a != null) { v.add(a.key); a = a.next; } while (b != null) { v.add(b.key); b = b.next; } Collections.sort(v); Node result = new Node(-1); Node temp = result; for (int i = 0; i < v.size(); i++) { result.next = new Node(v.get(i)); result = result.next; } temp = temp.next; System.out.print("Resultant Merge Linked List is : "); while (temp != null) { System.out.print(temp.key + " "); temp = temp.next; } }} |
Python3
# Python program for the above approach# link list nodeclass Node: def __init__(self, key): self.key = key self.next = None # return a newnodedef newNode(key): return Node(key)# driver code#let us create two sorted linked lists to test the above#function. Created lists shall be#a: 5->10->15->40#b: 2->3->20a = Node(5)a.next = Node(10)a.next.next = Node(15)a.next.next.next = Node(40)b = Node(2)b.next = Node(3)b.next.next = Node(20)v = []while(a is not None): v.append(a.key) a = a.nextwhile(b is not None): v.append(b.key) b = b.nextv.sort()result = Node(-1)temp = resultfor i in range(len(v)): result.next = Node(v[i]) result = result.nexttemp = temp.nextprint("Resultant Merge Linked List is : ")while(temp is not None): print(temp.key, end=" ") temp = temp.next# THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999) |
C#
using System;using System.Collections.Generic;using System.Collections;using System.Linq;/* Link list Node */public class Node { public int key; public Node next; public Node(int d) { key = d; next = null; }}// function to sortclass GFG : IComparer<int>{ public int Compare(int x, int y) { if (x == 0 || y == 0) { return 0; } // CompareTo() method return x.CompareTo(y); }}class HelloWorld { public static Node newNode(int key){ Node temp = new Node(key); temp.next = null; return temp; } static void Main() { /* Let us create two sorted linked lists to test the above functions. Created lists shall be a: 5->10->15->40 b: 2->3->20 */ Node a = newNode(5); a.next = newNode(10); a.next.next = newNode(15); a.next.next.next = newNode(40); Node b = newNode(2); b.next = newNode(3); b.next.next = newNode(20); List<int> v = new List<int>(); while(a!=null){ v.Add(a.key); a=a.next; } while(b!=null){ v.Add(b.key); b=b.next; } GFG gg = new GFG(); v.Sort(gg); Node result= newNode(-1); Node temp=result; for(int i=0;i<v.Count;i++){ result.next=newNode(v[i]); result=result.next; } temp=temp.next; Console.WriteLine("Resultant Merge Linked List Is :"); while(temp!=null){ Console.Write(temp.key + " "); temp=temp.next; } }}// The code is contributed by Nidhi goel. |
Javascript
// JavaScript program for the above approach// Link List nodeclass Node{ constructor(key){ this.key = key; this.next = null; }}function newNode(key){ let temp = new Node(key); return temp;}// driver code// let us create two sorted linked lists to test the above// function. Created lists shall be// a: 5->10->15->40// b: 2->3->20let a = newNode(5);a.next = newNode(10);a.next.next = newNode(15);a.next.next.next = newNode(40);let b = newNode(2);b.next = newNode(3);b.next.next = newNode(20);let v = [];while(a != null){ v.push(a.key); a = a.next;}while(b != null){ v.push(b.key); b = b.next;}v.sort(function(a, b){return a - b});let result = newNode(-1);let temp = result;for(let i = 0; i<v.length; i++){ result.next = newNode(v[i]); result = result.next;}temp = temp.next;console.log("Resultant Merge Linked List is : ");while(temp != null){ console.log(temp.key + " "); temp = temp.next;}// This code is contributed by Yash Agarwal(yashagarwal2852002) |
Output
Resultant Merge Linked List Is : 2 3 5 10 15 20 40
Complexity Analysis:
Time Complexity:O((n+m)*(n+m)log(n+m)),(n+m)for traversing linked lists and (n+m)log(n+m) for sorting vector.
Auxiliary Space: O(n+m),for vector.
Merge two sorted linked lists using Dummy Nodes:
The idea is to use 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.
Follow the below illustration for a better understanding:
Illustration:
Follow the steps below to solve the problem:
- First, make a dummy node for the new merged linked list
- Now make two pointers, one will point to list1 and another will point to list2.
- Now traverse the lists till one of them gets exhausted.
- If the value of the node pointing to either list is smaller than another pointer, add that node to our merged list and increment that pointer.
Below is the implementation of the above approach:
C++
/* C++ program to merge two sorted linked lists */#include <bits/stdc++.h>using namespace std;/* Link list node */class Node {public: int data; Node* next;};/* pull of the front node ofthe source and put it in dest */void MoveNode(Node** destRef, Node** sourceRef);/* Takes two lists sorted in increasingorder, and splices their nodes togetherto make one big sorted list whichis returned. */Node* SortedMerge(Node* a, Node* b){ /* a dummy first node to hang the result on */ Node dummy; /* tail points to the last result node */ 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 thenode from the front of the source,and move it to the front of the dest.It is an error to call this with thesource list empty.Before calling MoveNode():source == {1, 2, 3}dest == {1, 2, 3}After calling MoveNode():source == {2, 3}dest == {1, 1, 2, 3} */void MoveNode(Node** destRef, Node** sourceRef){ /* the front source node */ Node* newNode = *sourceRef; assert(newNode != NULL); /* Advance the source pointer */ *sourceRef = newNode->next; /* Link the old dest of the new node */ newNode->next = *destRef; /* Move dest to point to the new node */ *destRef = newNode;}/* Function to insert a node atthe beginning of the linked list */void push(Node** head_ref, int new_data){ /* allocate node */ Node* new_node = new Node(); /* put in the data */ new_node->data = new_data; /* link the old list of 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(Node* node){ while (node != NULL) { cout << node->data << " "; node = node->next; }}/* Driver code*/int main(){ /* Start with the empty list */ Node* res = NULL; Node* a = NULL; 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); cout << "Merged Linked List is: \n"; printList(res); return 0;}// This code is contributed by rathbhupendra |
C
/* C program to merge two sorted linked lists */#include <assert.h>#include <stdio.h>#include <stdlib.h>/* Link list node */struct Node { int data; struct Node* next;};/* pull of 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} After 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 of the new node */ newNode->next = *destRef; /* Move dest to point to the new node */ *destRef = newNode;}/* Function to insert a node at the beginning 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 of 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; }}/* Driver 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); System.out.println("Merged Linked List is:"); 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 |
Python3
""" Python program to merge twosorted linked lists """# Linked List Nodeclass Node: def __init__(self, data): self.data = data self.next = None# Create & Handle List operationsclass LinkedList: def __init__(self): self.head = None # Method to display the list def printList(self): temp = self.head while temp: print(temp.data, end=" ") temp = temp.next # Method to add element to list def addToList(self, newData): newNode = Node(newData) if self.head is None: self.head = newNode return last = self.head while last.next: last = last.next last.next = newNode# Function to merge the lists# Takes two lists which are sorted# joins them to get a single sorted listdef mergeLists(headA, headB): # A dummy node to store the result dummyNode = Node(0) # Tail stores the last node tail = dummyNode while True: # If any of the list gets completely empty # directly join all the elements of the other list if headA is None: tail.next = headB break if headB is None: tail.next = headA break # Compare the data of the lists and whichever is smaller is # appended to the last of the merged list and the head is changed if headA.data <= headB.data: tail.next = headA headA = headA.next else: tail.next = headB headB = headB.next # Advance the tail tail = tail.next # Returns the head of the merged list return dummyNode.next# Create 2 listslistA = LinkedList()listB = LinkedList()# Add elements to the list in sorted orderlistA.addToList(5)listA.addToList(10)listA.addToList(15)listB.addToList(2)listB.addToList(3)listB.addToList(20)# Call the merge functionlistA.head = mergeLists(listA.head, listB.head)# Display merged listprint("Merged Linked List is:")listA.printList()""" This code is contributedby Debidutta Rath """ |
C#
/* C# program to merge twosorted linked lists */using System;/* Link list node */public class Node { public int data; public Node next; public Node(int d) { data = d; next = null; }}public 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) { Console.Write(temp.data + " "); temp = temp.next; } Console.WriteLine(); } // 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); Console.WriteLine("Merged Linked List is:"); llist1.printList(); }}public class Gfg { /* Takes two lists sorted in increasing order, and splices their nodes together to make one big sorted list which is returned. */ public 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 29AjayKumar |
Javascript
<script>class Node{ constructor(d) { this.data=d; this.next = null; }}class LinkedList{constructor(){ this.head=null;}addToTheLast(node){ if (this.head == null) { this.head = node; } else { let temp = this.head; while (temp.next != null) temp = temp.next; temp.next = node; }}printList(){ let temp = this.head; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } document.write("<br>");}}function sortedMerge(headA,headB){ /* a dummy first node to hang the result on */ let dummyNode = new Node(0); /* tail points to the last result node */ let 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;}/* Let us create two sorted linked lists to test the methods Created lists: llist1: 5->10->15, llist2: 2->3->20 */let llist1 = new LinkedList();let llist2 = new LinkedList();// 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 = sortedMerge(llist1.head, llist2.head);document.write("Merged Linked List is:<br>")llist1.printList(); // This code is contributed by patel2127</script> |
Output
Merged Linked List is: 2 3 5 10 15 20
Time Complexity: O(N + M), where N and M are the size of list1 and list2 respectively
Auxiliary Space: O(1)
Merge two sorted linked lists using Recursion:
The idea is to move ahead with node in the recursion whose node value is lesser. When any of the node reach the end then append the rest of the linked List.
Follow the steps below to solve the problem:
- Make a function where two pointers pointing to the linked list will be passed.
- Now, check which value is less from both the current nodes
- The one with less value makes a recursion call by moving ahead with that pointer and simultaneously append that recursion call with the node
- Also put two base cases to check whether one of the linked lists will reach the NULL, then append the rest of the linked list.
Below is the implementation of the above approach.
C++
/* C++ program to merge two sorted linked lists */#include <bits/stdc++.h>using namespace std;/* Link list node */class Node {public: int data; Node* next;};/* Takes two lists sorted in increasingorder, and splices their nodes togetherto make one big sorted list whichis returned. */Node* SortedMerge(Node* a, Node* b){ 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);}/* Function to insert a node atthe beginning of the linked list */void push(Node** head_ref, int new_data){ /* allocate node */ Node* new_node = new Node(); /* put in the data */ new_node->data = new_data; /* link the old list of 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(Node* node){ while (node != NULL) { cout << node->data << " "; node = node->next; }}/* Driver code*/int main(){ /* Start with the empty list */ Node* res = NULL; Node* a = NULL; 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); cout << "Merged Linked List is: \n"; printList(res); return 0;}// This code is contributed by rathbhupendra |
C
/* C program to merge two sorted linked lists */#include <assert.h>#include <stdio.h>#include <stdlib.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){ 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);}/* 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} After 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 beginning 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 of 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; }}/* Driver 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() { System.out.println("Merged Linked List is:"); 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. */ public Node sortedMerge(Node A, Node B) { if (A == null) return B; if (B == null) return A; if (A.data < B.data) { A.next = sortedMerge(A.next, B); return A; } else { B.next = sortedMerge(A, B.next); return B; } }}// This code is contributed by Tuhin Das |
Python3
# Python3 program merge two sorted linked# in third linked list using recursive.# Node classclass Node: def __init__(self, data): self.data = data self.next = None# Constructor to initialize the node objectclass 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 Functionif __name__ == '__main__': # Create linked list : list1 = LinkedList() list1.append(5) list1.append(10) list1.append(15) # Create linked list 2 : list2 = LinkedList() list2.append(2) list2.append(3) list2.append(20) # 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:") list3.printList()# This code is contributed by 'Shriaknt13'. |
C#
/* C# program to merge twosorted linked lists */using System;/* Link list node */public class Node { public int data; public Node next; public Node(int d) { data = d; next = null; }}public 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() { Console.WriteLine("Merged Linked List is:"); Node temp = head; while (temp != null) { Console.Write(temp.data + " "); temp = temp.next; } Console.WriteLine(); } // 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(); }}public class Gfg { /* Takes two lists sorted in increasing order, and splices their nodes together to make one big sorted list which is returned. */ public Node sortedMerge(Node A, Node B) { // Base cases if (A == null) return B; if (B == null) return A; // Pick either a or b, and recur if (A.data < B.data) { A.next = sortedMerge(A.next, B); return A; } else { B.next = sortedMerge(A, B.next); return B; } }} |
Javascript
<script> function SortedMerge( A, B) { if (A == null) return B; if (B == null) return A; if (A.data < B.data) { A.next = SortedMerge(A.next, B); return A; } else { B.next = SortedMerge(A, B.next); return B; } }// This code contributed by umadevi9616</script> |
Output
Merged Linked List is: 2 3 5 10 15 20
Time Complexity: O(M + N), Where M and N are the size of the list1 and list2 respectively.
Auxiliary Space: O(M+N), Function call stack space
Merge two sorted linked lists by Reversing the Lists:
This idea involves first reversing both the given lists and after reversing, traversing both the lists till the end and then comparing the nodes of both the lists and inserting the node with a larger value at the beginning of the result list. And in this way, we will get the resulting list in increasing order.
Follow the steps below to solve the problem:
- Initialize result list as empty: head = NULL.
- Let ‘a’ and ‘b’ be the heads of the first and second lists respectively.
- Reverse both lists.
- While (a != NULL and b != NULL)
- Find the larger of two (Current ‘a’ and ‘b’)
- Insert the larger value of the node at the front of result list.
- Move ahead in the list with the larger node.
- If ‘b’ becomes NULL before ‘a’, insert all nodes of ‘a’ into result list at the beginning.
- If ‘a’ becomes NULL before ‘b’, insert all nodes of ‘b’ into result list at the beginning.
Below is the implementation of the above solution.
C++
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/#include <bits/stdc++.h>using namespace std;/* Link list Node */struct Node { int key; struct Node* next;};// Function to reverse a given Linked List using RecursionNode* reverseList(Node* head){ if (head->next == NULL) return head; Node* rest = reverseList(head->next); head->next->next = head; head->next = NULL; return rest;}// Given two non-empty linked lists 'a' and 'b'Node* sortedMerge(Node* a, Node* b){ // Reverse Linked List 'a' a = reverseList(a); // Reverse Linked List 'b' b = reverseList(b); // Initialize head of resultant list Node* head = NULL; Node* temp; // Traverse both lists while both of them // have nodes. while (a != NULL && b != NULL) { // If a's current value is greater than or equal to // b's current value. if (a->key >= b->key) { // Store next of current Node in first list temp = a->next; // Add 'a' at the front of resultant list a->next = head; // Make 'a' - head of the result list head = a; // Move ahead in first list a = temp; } // If b's value is greater. Below steps are similar // to above (Only 'a' is replaced with 'b') else { temp = b->next; b->next = head; head = b; b = temp; } } // If second list reached end, but first list has // nodes. Add remaining nodes of first list at the // beginning of result list while (a != NULL) { temp = a->next; a->next = head; head = a; a = temp; } // If first list reached end, but second list has // nodes. Add remaining nodes of second list at the // beginning of result list while (b != NULL) { temp = b->next; b->next = head; head = b; b = temp; } // Return the head of the result list return head;}/* Function to print Nodes in a given linked list */void printList(struct Node* Node){ while (Node != NULL) { cout << Node->key << " "; Node = Node->next; }}/* Utility function to create a new node with given key */Node* newNode(int key){ Node* temp = new Node; temp->key = key; temp->next = NULL; return temp;}/* Driver program to test above functions*/int main(){ /* Start with the empty list */ struct Node* res = NULL; /* Let us create two sorted linked lists to test the above functions. Created lists shall be a: 5->10->15->40 b: 2->3->20 */ Node* a = newNode(5); a->next = newNode(10); a->next->next = newNode(15); a->next->next->next = newNode(40); Node* b = newNode(2); b->next = newNode(3); b->next->next = newNode(20); /* merge 2 sorted Linked Lists */ res = sortedMerge(a, b); cout << "Merged Linked List is:" << endl; printList(res); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/#include <stdio.h>#include <stdlib.h>/* Link list Node */typedef struct Node { int key; struct Node* next;} Node;// Function to reverse a given Linked List using RecursionNode* reverseList(Node* head){ if (head->next == NULL) return head; Node* rest = reverseList(head->next); head->next->next = head; head->next = NULL; return rest;}// Given two non-empty linked lists 'a' and 'b'Node* sortedMerge(Node* a, Node* b){ // Reverse Linked List 'a' a = reverseList(a); // Reverse Linked List 'b' b = reverseList(b); // Initialize head of resultant list Node* head = NULL; Node* temp; // Traverse both lists while both of them // have nodes. while (a != NULL && b != NULL) { // If a's current value is greater than or equal to // b's current value. if (a->key >= b->key) { // Store next of current Node in first list temp = a->next; // Add 'a' at the front of resultant list a->next = head; // Make 'a' - head of the result list head = a; // Move ahead in first list a = temp; } // If b's value is greater. Below steps are similar // to above (Only 'a' is replaced with 'b') else { temp = b->next; b->next = head; head = b; b = temp; } } // If second list reached end, but first list has // nodes. Add remaining nodes of first list at the // beginning of result list while (a != NULL) { temp = a->next; a->next = head; head = a; a = temp; } // If first list reached end, but second list has // nodes. Add remaining nodes of second list at the // beginning of result list while (b != NULL) { temp = b->next; b->next = head; head = b; b = temp; } // Return the head of the result list return head;}/* Function to print Nodes in a given linked list */void printList(struct Node* Node){ while (Node != NULL) { printf("%d ", Node->key); Node = Node->next; }}/* Utility function to create a new node with given key */Node* newNode(int key){ Node* temp = (Node*)malloc(sizeof(Node)); temp->key = key; temp->next = NULL; return temp;}/* Driver program to test above functions*/int main(){ /* Start with the empty list */ struct Node* res = NULL; /* Let us create two sorted linked lists to test the above functions. Created lists shall be a: 5->10->15->40 b: 2->3->20 */ Node* a = newNode(5); a->next = newNode(10); a->next->next = newNode(15); a->next->next->next = newNode(40); Node* b = newNode(2); b->next = newNode(3); b->next->next = newNode(20); /* merge 2 sorted Linked Lists */ res = sortedMerge(a, b); printf("Merged Linked List is: \n"); printList(res); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/import java.util.*;class GFG { /* Link list Node */ static class Node { int key; Node next; }; // Function to reverse a given Linked List using // Recursion static Node reverseList(Node head) { if (head.next == null) return head; Node rest = reverseList(head.next); head.next.next = head; head.next = null; return rest; } // Given two non-empty linked lists 'a' and 'b' static Node sortedMerge(Node a, Node b) { // Reverse Linked List 'a' a = reverseList(a); // Reverse Linked List 'b' b = reverseList(b); // Initialize head of resultant list Node head = null; Node temp; // Traverse both lists while both of them // have nodes. while (a != null && b != null) { // If a's current value is greater than or equal // to b's current value. if (a.key >= b.key) { // Store next of current Node in first list temp = a.next; // Add 'a' at the front of resultant list a.next = head; // Make 'a' - head of the result list head = a; // Move ahead in first list a = temp; } // If b's value is greater. Below steps are // similar to above (Only 'a' is replaced with // 'b') else { temp = b.next; b.next = head; head = b; b = temp; } } // If second list reached end, but first list has // nodes. Add remaining nodes of first list at the // beginning of result list while (a != null) { temp = a.next; a.next = head; head = a; a = temp; } // If first list reached end, but second list has // nodes. Add remaining nodes of second list at the // beginning of result list while (b != null) { temp = b.next; b.next = head; head = b; b = temp; } // Return the head of the result list return head; } /* Function to print Nodes in a given linked list */ static void printList(Node Node) { while (Node != null) { System.out.print(Node.key + " "); Node = Node.next; } } /* Utility function to create a new node with given key */ static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.next = null; return temp; } /* Driver program to test above functions*/ public static void main(String[] args) { /* Start with the empty list */ Node res = null; /* Let us create two sorted linked lists to test the above functions. Created lists shall be a: 5.10.15.40 b: 2.3.20 */ Node a = newNode(5); a.next = newNode(10); a.next.next = newNode(15); a.next.next.next = newNode(40); Node b = newNode(2); b.next = newNode(3); b.next.next = newNode(20); /* merge 2 sorted Linked Lists */ res = sortedMerge(a, b); System.out.println("Merged Linked List is:"); printList(res); }}// This code is contributed by umadevi9616 |
Python3
# Given two sorted linked lists consisting of N and M nodes# respectively. The task is to merge both of the list# (in-place) and return head of the merged list.# Link list Nodeclass Node: def __init__(self, key): self.key = key self.next = None# Function to reverse a given Linked List using# Recursiondef reverse_list(head): if head.next is None: return head rest = reverse_list(head.next) head.next.next = head head.next = None return rest# Given two non-empty linked lists 'a' and 'b'def sorted_merge(a, b): # Reverse Linked List 'a' a = reverse_list(a) # Reverse Linked List 'b' b = reverse_list(b) # Initialize head of resultant list head = None # Traverse both lists while both of them # have nodes. while a is not None and b is not None: # If a's current value is greater than or equal # to b's current value. if a.key >= b.key: # Store next of current Node in first list temp = a.next # Add 'a' at the front of resultant list a.next = head # Make 'a' - head of the result list head = a # Move ahead in first list a = temp # If b's value is greater. Below steps are # similar to above (Only 'a' is replaced with # 'b') else: temp = b.next b.next = head head = b b = temp # If second list reached end, but first list has # nodes. Add remaining nodes of first list at the # beginning of result list while a is not None: temp = a.next a.next = head head = a a = temp # If first list reached end, but second list has # nodes. Add remaining nodes of second list at the # beginning of result list while b is not None: temp = b.next b.next = head head = b b = temp # Return the head of the result list return head# Function to print Nodes in a given linked listdef print_list(node): while node is not None: print(node.key, end=" ") node = node.next print()# Utility function to create a new node with# given keydef new_node(key): temp = Node(key) return temp# Test# Start with the empty listres = None# Let us create two sorted linked lists to test# the above functions. Created lists shall be# a: 5.10.15.40# b: 2.3.20a = new_node(5)a.next = new_node(10)a.next.next = new_node(15)a.next.next.next = new_node(40)b = new_node(2)b.next = new_node(3)b.next.next = new_node(20)# merge 2 sorted linked listsres = sorted_merge(a, b)print("Merged Linked List is:")print_list(res)# This code is contributed by lokesh |
C#
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/using System;using System.Collections.Generic;public class GFG { /* Link list Node */ public class Node { public int key; public Node next; }; // Function to reverse a given Linked List using // Recursion static Node reverseList(Node head) { if (head.next == null) return head; Node rest = reverseList(head.next); head.next.next = head; head.next = null; return rest; } // Given two non-empty linked lists 'a' and 'b' static Node sortedMerge(Node a, Node b) { // Reverse Linked List 'a' a = reverseList(a); // Reverse Linked List 'b' b = reverseList(b); // Initialize head of resultant list Node head = null; Node temp; // Traverse both lists while both of them // have nodes. while (a != null && b != null) { // If a's current value is greater than or equal // to b's current value. if (a.key >= b.key) { // Store next of current Node in first list temp = a.next; // Add 'a' at the front of resultant list a.next = head; // Make 'a' - head of the result list head = a; // Move ahead in first list a = temp; } // If b's value is greater. Below steps are // similar to above (Only 'a' is replaced with // 'b') else { temp = b.next; b.next = head; head = b; b = temp; } } // If second list reached end, but first list has // nodes. Add remaining nodes of first list at the // beginning of result list while (a != null) { temp = a.next; a.next = head; head = a; a = temp; } // If first list reached end, but second list has // nodes. Add remaining nodes of second list at the // beginning of result list while (b != null) { temp = b.next; b.next = head; head = b; b = temp; } // Return the head of the result list return head; } /* Function to print Nodes in a given linked list */ static void printList(Node Node) { while (Node != null) { Console.Write(Node.key + " "); Node = Node.next; } } /* Utility function to create a new node with given key */ static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.next = null; return temp; } /* Driver program to test above functions*/ public static void Main(String[] args) { /* Start with the empty list */ Node res = null; /* Let us create two sorted linked lists to test the above functions. Created lists shall be a: 5.10.15.40 b: 2.3.20 */ Node a = newNode(5); a.next = newNode(10); a.next.next = newNode(15); a.next.next.next = newNode(40); Node b = newNode(2); b.next = newNode(3); b.next.next = newNode(20); /* merge 2 sorted Linked Lists */ res = sortedMerge(a, b); Console.WriteLine("Merged Linked List is:"); printList(res); }}// This code is contributed by umadevi9616 |
Javascript
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/class Node { constructor(key) { this.key = key; this.next = null; }}// Function to reverse a given linked list using recursionfunction reverseList(head) { if (!head.next) { return head; } const rest = reverseList(head.next); head.next.next = head; head.next = null; return rest;}// Given two non-empty linked lists 'a' and 'b'function sortedMerge(a, b) { // Reverse linked list 'a' a = reverseList(a); // Reverse linked list 'b' b = reverseList(b); // Initialize head of resultant list let head = null; let temp; // Traverse both lists while both of them have nodes while (a && b) { // If a's current value is greater than or equal to b's current value if (a.key >= b.key) { // Store next of current node in first list temp = a.next; // Add 'a' at the front of resultant list a.next = head; // Make 'a' the head of the result list head = a; // Move ahead in first list a = temp; } // If b's value is greater. Below steps are similar to above (only 'a' is replaced with 'b') else { temp = b.next; b.next = head; head = b; b = temp; } } // If second list reached end, but first list has nodes. // Add remaining nodes of first list at the beginning of result list while (a) { temp = a.next; a.next = head; head = a; a = temp; } // If first list reached end, but second list has nodes. // Add remaining nodes of second list at the beginning of result list while (b) { temp = b.next; b.next = head; head = b; b = temp; } // Return the head of the result list return head;}// Function to print nodes in a given linked listfunction printList(node) { let result = ""; while (node) { result += node.key + " "; node = node.next; } console.log(result);}// Start with the empty listlet res = null;// Create two sorted linked lists to test the above functions// List a: 5 -> 10 -> 15 -> 40// List b: 2 -> 3 -> 20const a = new Node(5);a.next = new Node(10);a.next.next = new Node(15);a.next.next.next = new Node(40);const b = new Node(2);b.next = new Node(3);b.next.next = new Node(20);/* merge 2 sorted Linked Lists */res = sortedMerge(a, b);console.log("Merged Linked List is:" + "<br>");printList(res);// This code is contributed by lokeshmvs21. |
Output
Merged Linked List is: 2 3 5 10 15 20 40
Time Complexity: O(M+N) where M and N are the lengths of the two lists to be merged.
Auxiliary Space: O(M+N)
This method is contributed by Mehul Mathur(mathurmehul01).
Please refer below post for simpler implementations :
Merge two sorted lists (in-place)
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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