Friday, January 27, 2023

Login

Zeroplusfour

No Result

View All Result

No Result

View All Result

Zeroplusfour

No Result

View All Result

Home Code Solutions Hackerrank Algorithms

Merge two sorted linked lists – HackerRank Solution

Merge two sorted linked lists - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

by bhautik bhalala

Reading Time: 1 min read

15 Days to learn SQL Hard SQL(Advanced)-Solution

15 Days to learn SQL Hard SQL(Advanced)-Solution alt text

Spread the love

Table of Contents

Merge two sorted linked lists – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Merge two sorted linked lists HackerRank Solution


#include <bits/stdc++.h>

using namespace std;

class SinglyLinkedListNode {
    public:
        int data;
        SinglyLinkedListNode *next;

        SinglyLinkedListNode(int node_data) {
            this->data = node_data;
            this->next = nullptr;
        }
};

class SinglyLinkedList {
    public:
        SinglyLinkedListNode *head;
        SinglyLinkedListNode *tail;

        SinglyLinkedList() {
            this->head = nullptr;
            this->tail = nullptr;
        }

        void insert_node(int node_data) {
            SinglyLinkedListNode* node = new SinglyLinkedListNode(node_data);

            if (!this->head) {
                this->head = node;
            } else {
                this->tail->next = node;
            }

            this->tail = node;
        }
};

void print_singly_linked_list(SinglyLinkedListNode* node, string sep, ofstream& fout) {
    while (node) {
        fout << node->data;

        node = node->next;

        if (node) {
            fout << sep;
        }
    }
}

void free_singly_linked_list(SinglyLinkedListNode* node) {
    while (node) {
        SinglyLinkedListNode* temp = node;
        node = node->next;

        free(temp);
    }
}


SinglyLinkedListNode* mergeLists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
    if(!head1) return head2;
    if(!head2) return head1;
    SinglyLinkedListNode *p1 = head1, *p2 = head2, *h, *p;
    if(p1->data < p2->data) {
        h = p1;
        p1 = p1->next;
    } else {
        h = p2;
        p2 = p2->next;
    }
    p = h;
    while(p1!=NULL || p2!=NULL) {
        if(!p2 || (p1 && p1->data < p2->data)) {
            p->next = p1;
            p1 = p1->next;
        } else {
            p->next = p2;
            p2 = p2->next;
        }
        p = p->next;
    }
    return h;
}


int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    int tests;
    cin >> tests;
    cin.ignore(numeric_limits<streamsize>::max(), '\n');

    for (int tests_itr = 0; tests_itr < tests; tests_itr++) {
        SinglyLinkedList* llist1 = new SinglyLinkedList();

        int llist1_count;
        cin >> llist1_count;
        cin.ignore(numeric_limits<streamsize>::max(), '\n');

        for (int i = 0; i < llist1_count; i++) {
            int llist1_item;
            cin >> llist1_item;
            cin.ignore(numeric_limits<streamsize>::max(), '\n');

            llist1->insert_node(llist1_item);
        }
      
        SinglyLinkedList* llist2 = new SinglyLinkedList();

        int llist2_count;
        cin >> llist2_count;
        cin.ignore(numeric_limits<streamsize>::max(), '\n');

        for (int i = 0; i < llist2_count; i++) {
            int llist2_item;
            cin >> llist2_item;
            cin.ignore(numeric_limits<streamsize>::max(), '\n');

            llist2->insert_node(llist2_item);
        }

        SinglyLinkedListNode* llist3 = mergeLists(llist1->head, llist2->head);

        print_singly_linked_list(llist3, " ", fout);
        fout << "\n";

        free_singly_linked_list(llist3);
    }

    fout.close();

    return 0;
}

Java Merge two sorted linked lists HackerRank Solution


import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    static class SinglyLinkedListNode {
        public int data;
        public SinglyLinkedListNode next;

        public SinglyLinkedListNode(int nodeData) {
            this.data = nodeData;
            this.next = null;
        }
    }

    static class SinglyLinkedList {
        public SinglyLinkedListNode head;
        public SinglyLinkedListNode tail;

        public SinglyLinkedList() {
            this.head = null;
            this.tail = null;
        }

        public void insertNode(int nodeData) {
            SinglyLinkedListNode node = new SinglyLinkedListNode(nodeData);

            if (this.head == null) {
                this.head = node;
            } else {
                this.tail.next = node;
            }

            this.tail = node;
        }
    }

    public static void printSinglyLinkedList(SinglyLinkedListNode node, String sep, BufferedWriter bufferedWriter) throws IOException {
        while (node != null) {
            bufferedWriter.write(String.valueOf(node.data));

            node = node.next;

            if (node != null) {
                bufferedWriter.write(sep);
            }
        }
    }

    
    static SinglyLinkedListNode mergeLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) { 
        if(head1==null) {
            return head2; 
        } 
        if(head2 == null) {
            return head1;
        }
       SinglyLinkedListNode t1 = head1, t2 = head2; 
       SinglyLinkedListNode head = null, tail = null;  
       if(t1.data<=t2.data) {
           head = t1; 
           tail = t1; 
           t1= t1.next;
       } else {
           head = t2; 
           tail = t2; 
           t2 =  t2.next;
       } 
       while(t1!=null && t2!=null) { 
           if(t1.data<=t2.data) {
               tail.next = t1;  
               tail = t1;
               t1 = t1.next; 
           } else {
               tail.next = t2;  
               tail = t2;
               t2 = t2.next;
           }

       } 
       if(t1!=null) {
           tail.next = t1; 
       } else {
           tail.next = t2; 
       } 
       return head; 
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int tests = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        for (int testsItr = 0; testsItr < tests; testsItr++) {
            SinglyLinkedList llist1 = new SinglyLinkedList();

            int llist1Count = scanner.nextInt();
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            for (int i = 0; i < llist1Count; i++) {
                int llist1Item = scanner.nextInt();
                scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

                llist1.insertNode(llist1Item);
            }
          
            SinglyLinkedList llist2 = new SinglyLinkedList();

            int llist2Count = scanner.nextInt();
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            for (int i = 0; i < llist2Count; i++) {
                int llist2Item = scanner.nextInt();
                scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

                llist2.insertNode(llist2Item);
            }

            SinglyLinkedListNode llist3 = mergeLists(llist1.head, llist2.head);

            printSinglyLinkedList(llist3, " ", bufferedWriter);
            bufferedWriter.newLine();
        }

        bufferedWriter.close();

        scanner.close();
    }
}

Python 3 Merge two sorted linked lists HackerRank Solution


#!/bin/python3

import math
import os
import random
import re
import sys

class SinglyLinkedListNode:
    def __init__(self, node_data):
        self.data = node_data
        self.next = None

class SinglyLinkedList:
    def __init__(self):
        self.head = None
        self.tail = None

    def insert_node(self, node_data):
        node = SinglyLinkedListNode(node_data)

        if not self.head:
            self.head = node
        else:
            self.tail.next = node


        self.tail = node

def print_singly_linked_list(node, sep, fptr):
    while node:
        fptr.write(str(node.data))

        node = node.next

        if node:
            fptr.write(sep)


def mergeLists(headA, headB):
    if headA == None:
        return headB
    if headB == None:
        return headA
    
    if headA.data <= headB.data:
        head = headA
        headA = headA.next
    else:
        head = headB
        headB = headB.next
    
    current = head
    while headA != None or headB != None:
        if headA == None:
            current.next = headB
            break
        if headB == None:
            current.next = headA
            break
        if headA.data <= headB.data:
            current.next = headA
            headA = headA.next
        else:
            current.next = headB
            headB = headB.next
        current = current.next
    return head

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    tests = int(input())

    for tests_itr in range(tests):
        llist1_count = int(input())

        llist1 = SinglyLinkedList()

        for _ in range(llist1_count):
            llist1_item = int(input())
            llist1.insert_node(llist1_item)
            
        llist2_count = int(input())

        llist2 = SinglyLinkedList()

        for _ in range(llist2_count):
            llist2_item = int(input())
            llist2.insert_node(llist2_item)

        llist3 = mergeLists(llist1.head, llist2.head)

        print_singly_linked_list(llist3, ' ', fptr)
        fptr.write('\n')

    fptr.close()

JavaScript Merge two sorted linked lists HackerRank Solution


'use strict';

const fs = require('fs');

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', inputStdin => {
    inputString += inputStdin;
});

process.stdin.on('end', _ => {
    inputString = inputString.replace(/\s*$/, '')
        .split('\n')
        .map(str => str.replace(/\s*$/, ''));

    main();
});

function readLine() {
    return inputString[currentLine++];
}

const SinglyLinkedListNode = class {
    constructor(nodeData) {
        this.data = nodeData;
        this.next = null;
    }
};

const SinglyLinkedList = class {
    constructor() {
        this.head = null;
        this.tail = null;
    }

    insertNode(nodeData) {
        const node = new SinglyLinkedListNode(nodeData);

        if (this.head == null) {
            this.head = node;
        } else {
            this.tail.next = node;
        }

        this.tail = node;
    }
};

function printSinglyLinkedList(node, sep, ws) {
    while (node != null) {
        ws.write(String(node.data));

        node = node.next;

        if (node != null) {
            ws.write(sep);
        }
    }
}


function mergeLists(headA, headB) {
    // Handle empty lists.
    if (!headA) return headB;
    if (!headB) return headA;
    
    var nodeA = headA;
    var nodeB = headB;
    
    var head = new SinglyLinkedListNode(null);
    var leader = head;
    while(nodeA || nodeB){
        // Handle reaching the end of either list.
        if (!nodeA){
            leader.next = nodeB;
            break;
        }
        if (!nodeB){
            leader.next = nodeA;
            break;
        }
        
        // Select the next lowest node.
        if(nodeA.data < nodeB.data){
            leader.next = nodeA;
            nodeA = nodeA.next;
        }
        else {
            leader.next = nodeB;
            nodeB = nodeB.next;
        }
        leader = leader.next;
    }
    
    return head.next;
}

function main() {
    const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

    const tests = parseInt(readLine(), 10);

    for (let testsItr = 0; testsItr < tests; testsItr++) {
        const llist1Count = parseInt(readLine(), 10);

        let llist1 = new SinglyLinkedList();

        for (let i = 0; i < llist1Count; i++) {
            const llist1Item = parseInt(readLine(), 10);
            llist1.insertNode(llist1Item);
        }
      
        const llist2Count = parseInt(readLine(), 10);

        let llist2 = new SinglyLinkedList();

        for (let i = 0; i < llist2Count; i++) {
            const llist2Item = parseInt(readLine(), 10);
            llist2.insertNode(llist2Item);
        }

        let llist3 = mergeLists(llist1.head, llist2.head);

        printSinglyLinkedList(llist3, " ", ws)
        ws.write("\n");
    }

    ws.end();
}

C Merge two sorted linked lists HackerRank Solution


#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* readline();

typedef struct SinglyLinkedListNode SinglyLinkedListNode;
typedef struct SinglyLinkedList SinglyLinkedList;

struct SinglyLinkedListNode {
    int data;
    SinglyLinkedListNode* next;
};

struct SinglyLinkedList {
    SinglyLinkedListNode* head;
    SinglyLinkedListNode* tail;
};

SinglyLinkedListNode* create_singly_linked_list_node(int node_data) {
    SinglyLinkedListNode* node = malloc(sizeof(SinglyLinkedListNode));

    node->data = node_data;
    node->next = NULL;

    return node;
}

void insert_node_into_singly_linked_list(SinglyLinkedList** singly_linked_list, int node_data) {
    SinglyLinkedListNode* node = create_singly_linked_list_node(node_data);

    if (!(*singly_linked_list)->head) {
        (*singly_linked_list)->head = node;
    } else {
        (*singly_linked_list)->tail->next = node;
    }

    (*singly_linked_list)->tail = node;
}

void print_singly_linked_list(SinglyLinkedListNode* node, char* sep, FILE* fptr) {
    while (node) {
        fprintf(fptr, "%d", node->data);

        node = node->next;

        if (node) {
            fprintf(fptr, "%s", sep);
        }
    }
}

void free_singly_linked_list(SinglyLinkedListNode* node) {
    while (node) {
        SinglyLinkedListNode* temp = node;
        node = node->next;

        free(temp);
    }
}


SinglyLinkedListNode* mergeLists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
    if (!head1) return head2;
    if (!head2) return head1;
    SinglyLinkedListNode* curr1 = head1;
    SinglyLinkedListNode* curr2 = head2;
    SinglyLinkedListNode* new_list;
    SinglyLinkedListNode* curr_new;
    if (head1->data < head2->data) {
        new_list = head1;
        curr1 = curr1->next;
    }
    else {
        new_list = head2;
        curr2 = curr2->next;
    }
    curr_new = new_list;
    while (curr1 && curr2) {
        if (curr1->data < curr2->data) {
            curr_new->next = curr1;
            
            curr1 = curr1->next;
        }
        else {
            curr_new->next = curr2;
            
            curr2 = curr2->next;
        }
        curr_new = curr_new->next;
    }
    if (curr1 != NULL) {
        curr_new->next = curr1;
    }
    if (curr2 != NULL) {
        curr_new->next = curr2;
    }
    return new_list;

}
int main()
{
    FILE* fptr = fopen(getenv("OUTPUT_PATH"), "w");

    char* tests_endptr;
    char* tests_str = readline();
    int tests = strtol(tests_str, &tests_endptr, 10);

    if (tests_endptr == tests_str || *tests_endptr != '\0') { exit(EXIT_FAILURE); }

    for (int tests_itr = 0; tests_itr < tests; tests_itr++) {
        SinglyLinkedList* llist1 = malloc(sizeof(SinglyLinkedList));
        llist1->head = NULL;
        llist1->tail = NULL;

        char* llist1_count_endptr;
        char* llist1_count_str = readline();
        int llist1_count = strtol(llist1_count_str, &llist1_count_endptr, 10);

        if (llist1_count_endptr == llist1_count_str || *llist1_count_endptr != '\0') { exit(EXIT_FAILURE); }

        for (int i = 0; i < llist1_count; i++) {
            char* llist1_item_endptr;
            char* llist1_item_str = readline();
            int llist1_item = strtol(llist1_item_str, &llist1_item_endptr, 10);

            if (llist1_item_endptr == llist1_item_str || *llist1_item_endptr != '\0') { exit(EXIT_FAILURE); }

            insert_node_into_singly_linked_list(&llist1, llist1_item);
        }
      
        SinglyLinkedList* llist2 = malloc(sizeof(SinglyLinkedList));
        llist2->head = NULL;
        llist2->tail = NULL;

        char* llist2_count_endptr;
        char* llist2_count_str = readline();
        int llist2_count = strtol(llist2_count_str, &llist2_count_endptr, 10);

        if (llist2_count_endptr == llist2_count_str || *llist2_count_endptr != '\0') { exit(EXIT_FAILURE); }

        for (int i = 0; i < llist2_count; i++) {
            char* llist2_item_endptr;
            char* llist2_item_str = readline();
            int llist2_item = strtol(llist2_item_str, &llist2_item_endptr, 10);

            if (llist2_item_endptr == llist2_item_str || *llist2_item_endptr != '\0') { exit(EXIT_FAILURE); }

            insert_node_into_singly_linked_list(&llist2, llist2_item);
        }

        SinglyLinkedListNode* llist3 = mergeLists(llist1->head, llist2->head);

        char *sep = " ";

        print_singly_linked_list(llist3, sep, fptr);
        fprintf(fptr, "\n");

        free_singly_linked_list(llist3);
    }

    fclose(fptr);

    return 0;
}

char* readline() {
    size_t alloc_length = 1024;
    size_t data_length = 0;
    char* data = malloc(alloc_length);

    while (true) {
        char* cursor = data + data_length;
        char* line = fgets(cursor, alloc_length - data_length, stdin);

        if (!line) { break; }

        data_length += strlen(cursor);

        if (data_length < alloc_length - 1 || data[data_length - 1] == '\n') { break; }

        size_t new_length = alloc_length << 1;
        data = realloc(data, new_length);

        if (!data) { break; }

        alloc_length = new_length;
    }

    if (data[data_length - 1] == '\n') {
        data[data_length - 1] = '\0';
    }

    data = realloc(data, data_length);

    return data;
}

Leave a comment below:

Tags: C c++14 full solution Go HackerRank Solution java java 15 java 7 java 8 java8 javascript Merge two sorted linked lists pypy 3 Python 2 python 3

bhautik bhalala

Related Posts

Leetcode All Problems Solutions

Code Solutions

Exclusive Time of Functions – LeetCode Solution

October 5, 2022

Exclusive Time of Functions - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions...

Read more

Leetcode All Problems Solutions

Smallest Range Covering Elements from K Lists – LeetCode Solution

October 5, 2022

Leetcode All Problems Solutions

Course Schedule III – LeetCode Solution

October 5, 2022

Leetcode All Problems Solutions

Maximum Product of Three Numbers – LeetCode Solution

September 11, 2022

Leetcode All Problems Solutions

Task Scheduler – LeetCode Solution

September 11, 2022

Leetcode All Problems Solutions

Valid Triangle Number – LeetCode Solution

September 11, 2022

Next Post

15 Days to learn SQL Hard SQL(Advanced)-Solution

Queue using Two Stacks - HackerRank Solution

15 Days to learn SQL Hard SQL(Advanced)-Solution

Balanced Brackets - HackerRank Solution

Leave a Reply Cancel reply

No Result

View All Result

Home
Category
- Business
- Culture
- Economy
- Lifestyle
- Health
- Travel
- Opinion
- Politics
- Tech
Landing Page
Support Forum
Contact Us

© 2022 ZeroPlusFour - Latest News & Blog.

Privacy Overview

This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may affect your browsing experience.

Necessary

Always Enabled

Necessary cookies are absolutely essential for the website to function properly. These cookies ensure basic functionalities and security features of the website, anonymously.

Cookie	Duration	Description
cookielawinfo-checkbox-analytics	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Analytics".
cookielawinfo-checkbox-functional	11 months	The cookie is set by GDPR cookie consent to record the user consent for the cookies in the category "Functional".
cookielawinfo-checkbox-necessary	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookies is used to store the user consent for the cookies in the category "Necessary".
cookielawinfo-checkbox-others	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Other.
cookielawinfo-checkbox-performance	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Performance".
viewed_cookie_policy	11 months	The cookie is set by the GDPR Cookie Consent plugin and is used to store whether or not user has consented to the use of cookies. It does not store any personal data.

Others

Are you sure want to unlock this post?

Unlock left : 0

Are you sure want to cancel subscription?