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Intersection of linked lists interviewbit solution Python

1. Using Hashing

The idea is to traverse the first list and store each node’s address in a hash table. Then traverse the second list and get the address of the first node present in the hash table. This node would be the intersection point. Following is the C++, Java, and Python implementation based on the idea:

C++

#include

using namespace std;

// A Linked List Node

struct Node

{

int data;

Node* next;

};

// Utility function to create a new node with the given data and

// pushes it onto the list's front

void push[Node*& headRef, int data]

{

// create a new linked list node from the heap

Node* newNode = new Node;

newNode->data = data;

newNode->next = headRef;

headRef = newNode;

}

// Function to find the intersection point of two linked lists using hashing

Node* findIntersection[Node* first, Node* second]

{

// maintain a set to store list nodes

unordered_set nodes;

// traverse the first list and insert the address of each node into the set

while [first]

{

nodes.insert[first];

first = first->next;

}

// now traverse the second list and find the first node that is

// already present in the set

while [second]

{

// return the current node if it is found in the set

if [nodes.find[second] != nodes.end[]] {

return second;

}

second = second->next;

}

// we reach here if lists do not intersect

return nullptr;

}

int main[]

{

// construct the first linked list [1 —> 2 —> 3 —> 4 —> 5 —> null]

Node* first = nullptr;

for [int i = 5; i > 0; i--] {

push[first, i];

}

// construct the second linked list [1 —> 2 —> 3 —> null]

Node* second = nullptr;

for [int i = 3; i > 0; i--] {

push[second, i];

}

// link tail of the second list to the fourth node of the first list

second->next->next->next = first->next->next->next;

Node* addr = findIntersection[first, second];

if [addr] {

cout 5 -> 6 ^ | 10 -> 2 -> 3 -> 4 Output: 14

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisites: Write a function to get the intersection point of two Linked Lists

Approach: Take two pointers for the heads of both the linked lists. If one of them reaches the end earlier then use it by moving it to the beginning of the other list. Once both of them go through reassigning they will be equidistance from the collision point.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach

#include

using namespace std;

/* Link list node */

class Node {

public:

int data;

Node* next;

};

// Function to return the intersection point

// of the two linked lists head1 and head2

int getIntesectionNode[Node* head1, Node* head2]

{

Node* current1 = head1;

Node* current2 = head2;

// If one of the head is NULL

if [!current1 or !current2]

return -1;

// Continue until we find intersection node

while [current1 and current2

and current1 != current2] {

current1 = current1->next;

current2 = current2->next;

// If we get intersection node

if [current1 == current2]

return current1->data;

// If one of them reaches end

if [!current1]

current1 = head2;

if [!current2]

current2 = head1;

}

return current1->data;

}

// Driver code

int main[]

{

Create two linked lists

1st 3->6->9->15->30

2nd 10->15->30

15 is the intersection point

Node* newNode;

// Addition of new nodes

Node* head1 = new Node[];

head1->data = 10;

Node* head2 = new Node[];

head2->data = 3;

newNode = new Node[];

newNode->data = 6;

head2->next = newNode;

newNode = new Node[];

newNode->data = 9;

head2->next->next = newNode;

newNode = new Node[];

newNode->data = 15;

head1->next = newNode;

head2->next->next->next = newNode;

newNode = new Node[];

newNode->data = 30;

head1->next->next = newNode;

head1->next->next->next = NULL;

cout 6->3 // represents number 563
List2: 8->4->2 // represents number 842
Output:
Resultant list: 1->4->0->5 // represents number 1405
Explanation: 563 + 842 = 1405

Input:
List1: 7->5->9->4->6 // represents number 75946
List2: 8->4 // represents number 84
Output:
Resultant list: 7->6->0->3->0// represents number 76030
Explanation: 75946+84=76030

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach: Traverse both lists to the end and add preceding zeros in the list with lesser digits. Then call a recursive function on the start nodes of both lists which calls itself for the next nodes of both lists till it gets to the end. This function creates a node for the sum of the current digits and returns the carry.

The steps are:

Traverse the two linked lists in order to add preceding zeros in case a list is having lesser digits than the other one.
Start from the head node of both lists and call a recursive function for the next nodes.
Continue it till the end of the lists.
Creates a node for current digits sum and returns the carry.

Below is the implementation of this approach.

C++

// C++ program to add two numbers

// represented by linked list

#include

using namespace std;

/* Linked list node */

class Node {

public:

int data;

Node* next;

};

/* Function to create a

new node with given data */

Node* newNode[int data]

{

Node* new_node = new Node[];

new_node->data = data;

new_node->next = NULL;

return new_node;

}

/* Function to insert a node at the

beginning of the Singly Linked List */

void push[Node** head_ref, int new_data]

{

/* allocate node */

Node* new_node = newNode[new_data];

/* link the old list off the new node */

new_node->next = [*head_ref];

/* move the head to point to the new node */

[*head_ref] = new_node;

}

/* Adds contents of two linked lists and

return the head node of resultant list */

Node* addTwoLists[Node* first, Node* second]

{

// res is head node of the resultant list

Node* res = NULL;

Node *temp, *prev = NULL;

int carry = 0, sum;

// while both lists exist

while [first != NULL

|| second != NULL] {

// Calculate value of next

// digit in resultant list.

// The next digit is sum of

// following things

// [i] Carry

// [ii] Next digit of first

// list [if there is a next digit]

// [ii] Next digit of second

// list [if there is a next digit]

sum = carry + [first ? first->data : 0]

+ [second ? second->data : 0];

// update carry for next calculation

carry = [sum >= 10] ? 1 : 0;

// update sum if it is greater than 10

sum = sum % 10;

// Create a new node with sum as data

temp = newNode[sum];

// if this is the first node then

// set it as head of the resultant list

if [res == NULL]

res = temp;

// If this is not the first

// node then connect it to the rest.

else

prev->next = temp;

// Set prev for next insertion

prev = temp;

// Move first and second

// pointers to next nodes

if [first]

first = first->next;

if [second]

second = second->next;

}

if [carry > 0]

temp->next = newNode[carry];

// return head of the resultant list

return res;

}

Node* reverse[Node* head]

{

if [head == NULL || head->next == NULL]

return head;

/* reverse the rest list and put

the first element at the end */

Node* rest = reverse[head->next];

head->next->next = head;

head->next = NULL;

/* fix the head pointer */

return rest;

}

// A utility function to print a linked list

void printList[Node* node]

{

while [node != NULL] {

cout data next;

}

cout 5->9->4->6

push[&first, 6];

push[&first, 4];

push[&first, 9];

push[&first, 5];

push[&first, 7];

printf["First List is "];

printList[first];

// create second list 8->4

push[&second, 4];

push[&second, 8];

cout next = NULL;

return new_node;

}

/* Function to insert a node

at the beginning of the Singly Linked List */

void push[struct Node** head_ref, int new_data]

{

/* allocate node */

struct Node* new_node = newNode[new_data];

/* link the old list off the new node */

new_node->next = [*head_ref];

/* move the head to point to the new node */

[*head_ref] = new_node;

}

/* Adds contents of two linked

lists and return the head node

of resultant list */

struct Node* addTwoLists[struct Node* first,

struct Node* second]

{

// res is head node of the resultant list

struct Node* res = NULL;

struct Node *temp, *prev = NULL;

int carry = 0, sum;

// while both lists exist

while [first != NULL || second != NULL] {

// Calculate value of next

// digit in resultant list.

// The next digit is sum

// of following things

// [i] Carry

// [ii] Next digit of first

// list [if there is a next digit]

// [ii] Next digit of second

// list [if there is a next digit]

sum = carry + [first ? first->data : 0]

+ [second ? second->data : 0];

// Update carry for next calculation

carry = [sum >= 10] ? 1 : 0;

// Update sum if it is greater than 10

sum = sum % 10;

// Create a new node with sum as data

temp = newNode[sum];

// if this is the first node then

// set it as head of the resultant list

if [res == NULL]

res = temp;

// If this is not the first node

// then connect it to the rest.

else

prev->next = temp;

// Set prev for next insertion

prev = temp;

// Move first and second

// pointers to next nodes

if [first]

first = first->next;

if [second]

second = second->next;

}

if [carry > 0]

temp->next = newNode[carry];

// return head of the resultant list

return res;

}

// A utility function to print a linked list

void printList[struct Node* node]

{

while [node != NULL] {

printf["%d ", node->data];

node = node->next;

}

printf["\n"];

}

/* Driver code */

int main[void]

{

struct Node* res = NULL;

struct Node* first = NULL;

struct Node* second = NULL;

// create first list 7->5->9->4->6

push[&first, 6];

push[&first, 4];

push[&first, 9];

push[&first, 5];

push[&first, 7];

printf["First List is "];

printList[first];

// create second list 8->4

push[&second, 4];

push[&second, 8];

printf["Second List is "];

printList[second];

// Add the two lists and see result

res = addTwoLists[first, second];

printf["Resultant list is "];

printList[res];

return 0;

}

Java

// Java program to add two numbers

// represented by linked list

class LinkedList {

static Node head1, head2;

static class Node {

int data;

Node next;

Node[int d] {

data = d;

next = null;

}

/* Adds contents of two linked lists and prints it */

void addTwoLists[Node first, Node second] {

Node start1 = new Node[0];

start1.next = first;

Node start2 = new Node[0];

start2.next = second;

addPrecedingZeros[start1, start2];

Node result = new Node[0];

if [sumTwoNodes[start1.next, start2.next, result] == 1] {

Node node = new Node[1];

node.next = result.next;

result.next = node;

}

printList[result.next];

}

/* Adds lists and returns the carry */

private int sumTwoNodes[Node first, Node second, Node result] {

if [first == null] {

return 0;

}

int number = first.data + second.data + sumTwoNodes[first.next, second.next, result];

Node node = new Node[number % 10];

node.next = result.next;

result.next = node;

return number / 10;

}

/* Appends preceding zeros in case a list is having lesser nodes than the other one*/

private void addPrecedingZeros[Node start1, Node start2] {

Node next1 = start1.next;

Node next2 = start2.next;

while [next1 != null && next2 != null] {

next1 = next1.next;

next2 = next2.next;

}

if [next1 == null && next2 != null] {

while [next2 != null] {

Node node = new Node[0];

node.next = start1.next;

start1.next = node;

next2 = next2.next;

}

} else if [next2 == null && next1 != null] {

while [next1 != null] {

Node node = new Node[0];

node.next = start2.next;

start2.next = node;

next1 = next1.next;

}

/* Utility function to print a linked list */

void printList[Node head] {

while [head != null] {

System.out.print[head.data + " "];

head = head.next;

}

System.out.println[""];

}

// Driver Code

public static void main[String[] args] {

LinkedList list = new LinkedList[];

// creating first list

list.head1 = new Node[7];

list.head1.next = new Node[5];

list.head1.next.next = new Node[9];

list.head1.next.next.next = new Node[4];

list.head1.next.next.next.next = new Node[6];

System.out.print["First List is "];

list.printList[head1];

// creating second list

list.head2 = new Node[8];

list.head2.next = new Node[4];

System.out.print["Second List is "];

list.printList[head2];

System.out.print["Resultant List is "];

// add the two lists and see the result

list.addTwoLists[head1, head2];

}

// this code is contributed by *Saurabh321Gupta*

}

Python3

# Python program to add two numbers represented by linked list

# Node class

class Node:

# Constructor to initialize the node object

def __init__[self, data]:

self.data = data

self.next = None

class LinkedList:

# Function to initialize head

def __init__[self]:

self.head = None

# Function to insert a new node at the beginning

def push[self, new_data]:

new_node = Node[new_data]

new_node.next = self.head

self.head = new_node

# Add contents of two linked lists and return the head

# node of resultant list

def addTwoLists[self, first, second]:

prev = None

temp = None

carry = 0

# While both list exists

while[first is not None or second is not None]:

# Calculate the value of next digit in

# resultant list

# The next digit is sum of following things

# [i] Carry

# [ii] Next digit of first list [if ther is a

# next digit]

# [iii] Next digit of second list [ if there

# is a next digit]

fdata = 0 if first is None else first.data

sdata = 0 if second is None else second.data

Sum = carry + fdata + sdata

# update carry for next calculation

carry = 1 if Sum >= 10 else 0

# update sum if it is greater than 10

Sum = Sum if Sum < 10 else Sum % 10

# Create a new node with sum as data

temp = Node[Sum]

# if this is the first node then set it as head

# of resultant list

if self.head is None:

self.head = temp

else:

prev.next = temp

# Set prev for next insertion

prev = temp

# Move first and second pointers to next nodes

if first is not None:

first = first.next

if second is not None:

second = second.next

if carry > 0:

temp.next = Node[carry]

# Utility function to print the linked LinkedList

def printList[self]:

temp = self.head

while[temp]:

print [temp.data,end=' ']

temp = temp.next

# Driver code

first = LinkedList[]

second = LinkedList[]

# Create first list

first.push[6]

first.push[4]

first.push[9]

first.push[5]

first.push[7]

print ["First List is ",end=' ']

first.printList[]

# Create second list

second.push[4]

second.push[8]

print ["\nSecond List is",end=' ']

second.printList[]

# Add the two lists and see result

res = LinkedList[]

res.addTwoLists[first.head, second.head]

print ["\nResultant list is",end=' ']

res.printList[]

# This code is contributed by Nikhil Kumar Singh[nickzuck_007]

// C# program to add two numbers

// represented by linked list

using System;

public class LinkedList {

Node head1, head2;

public class Node {

public int data;

public Node next;

public Node[int d]

{

data = d;

next = null;

}

/* Adds contents of two linked lists

and return the head node of resultant list */

Node addTwoLists[Node first, Node second]

{

// res is head node of the resultant list

Node res = null;

Node prev = null;

Node temp = null;

int carry = 0, sum;

// while both lists exist

while [first != null || second != null] {

// Calculate value of next digit in resultant

// list. The next digit is sum of following

// things [i] Carry [ii] Next digit of first

// list [if there is a next digit] [ii] Next

// digit of second list [if there is a next

// digit]

sum = carry + [first != null ? first.data : 0]

+ [second != null ? second.data : 0];

// update carry for next calculation

carry = [sum >= 10] ? 1 : 0;

// update sum if it is greater than 10

sum = sum % 10;

// Create a new node with sum as data

temp = new Node[sum];

// if this is the first node then set it as head

// of the resultant list

if [res == null] {

res = temp;

}

// If this is not the first

// node then connect it to the rest.

else {

prev.next = temp;

}

// Set prev for next insertion

prev = temp;

// Move first and second pointers to next nodes

if [first != null] {

first = first.next;

}

if [second != null] {

second = second.next;

}

if [carry > 0] {

temp.next = new Node[carry];

}

// return head of the resultant list

return res;

}

/* Utility function to print a linked list */

void printList[Node head]

{

while [head != null] {

Console.Write[head.data + " "];

head = head.next;

}

Console.WriteLine[""];

}

// Driver code

public static void Main[String[] args]

{

LinkedList list = new LinkedList[];

// creating first list

list.head1 = new Node[7];

list.head1.next = new Node[5];

list.head1.next.next = new Node[9];

list.head1.next.next.next = new Node[4];

list.head1.next.next.next.next = new Node[6];

Console.Write["First List is "];

list.printList[list.head1];

// creating second list

list.head2 = new Node[8];

list.head2.next = new Node[4];

Console.Write["Second List is "];

list.printList[list.head2];

// add the two lists and see the result

Node rs = list.addTwoLists[list.head1, list.head2];

Console.Write["Resultant List is "];

list.printList[rs];

}

// This code contributed by Rajput-Ji

Javascript

// Javascript program to add two numbers

// represented by linked list

var head1, head2;

class Node {

constructor[val] {

this.data = val;

this.next = null;

}

Adds contents of two linked lists and return

the head node of resultant list

function addTwoLists[ first, second] {

// res is head node of the resultant list

var res = null;

var prev = null;

var temp = null;

var carry = 0, sum;

// while both lists exist

while [first != null || second != null] {

// Calculate value of next

// digit in resultant list.

// The next digit is sum

// of following things

// [i] Carry

// [ii] Next digit of first

// list [if there is a next digit]

// [ii] Next digit of second

// list [if there is a next digit]

sum = carry + [first != null ? first.data : 0] +

[second != null ? second.data : 0];

// update carry for next calculation

carry = [sum >= 10] ? 1 : 0;

// update sum if it is greater than 10

sum = sum % 10;

// Create a new node with sum as data

temp = new Node[sum];

// if this is the first node then set

// it as head of the resultant list

if [res == null] {

res = temp;

}

// If this is not the first

// node then connect it to the rest.

else {

prev.next = temp;

}

// Set prev for next insertion

prev = temp;

// Move first and second pointers

// to next nodes

if [first != null] {

first = first.next;

}

if [second != null] {

second = second.next;

}

if [carry > 0] {

temp.next = new Node[carry];

}

// return head of the resultant list

return res;

}

/* Utility function to print a linked list */

function printList[ head] {

while [head != null] {

document.write[head.data + " "];

head = head.next;

}

document.write["
"];

}

// Driver Code

// creating first list

head1 = new Node[7];

head1.next = new Node[5];

head1.next.next = new Node[9];

head1.next.next.next = new Node[4];

head1.next.next.next.next = new Node[6];

document.write["First List is "];

printList[head1];

// creating second list

head2 = new Node[8];

head2.next = new Node[4];

document.write["Second List is "];

printList[head2];

// add the two lists and see the result

rs = addTwoLists[head1, head2];

document.write["Resultant List is "];

printList[rs];

// This code contributed by aashish2995

Output First List is 7 5 9 4 6 Second List is 8 4 Resultant list is 5 0 0 5 6

Complexity Analysis:

Time Complexity: O[m + n], where m and n are numbers of nodes in first and second lists respectively.
The lists need to be traversed only once.
Space Complexity: O[m + n].
A temporary linked list is needed to store the output number

Method 2[Using STL]: Using the stack data structure

Approach :

Create 3 stacks namely s1,s2,s3.

Fill s1 with Nodes of list1 and fill s2 with nodes of list2.

Fill s3 by creating new nodes and setting the data of new nodes to the sum of s1.top[], s2.top[] and carry until list1 and list2 are empty .

If the sum >9

set carry 1

else

set carry 0

Create a Node[say prev] that will contain the head of the sum List.

Link all the elements of s3 from top to bottom

return prev

Code:

C++

// C++ program to add two numbers represented by Linked

// Lists using Stack

#include

using namespace std;

class Node {

public:

int data;

Node* next;

};

Node* newnode[int data]

{

Node* x = new Node[];

x->data = data;

return x;

}

// function that returns the sum of two numbers represented

// by linked lists

Node* addTwoNumbers[Node* l1, Node* l2]

{

Node* prev = NULL;

// Create 3 stacks

stack s1, s2, s3;

// Fill first stack with first List Elements

while [l1 != NULL] {

s1.push[l1];

l1 = l1->next;

}

// Fill second stack with second List Elements

while [l2 != NULL] {

s2.push[l2];

l2 = l2->next;

}

int carry = 0;

// Fill the third stack with the sum of first and second

// stack

while [!s1.empty[] && !s2.empty[]] {

int sum = s1.top[]->data + s2.top[]->data + carry;

Node* temp = newnode[sum % 10];

s3.push[temp];

if [sum > 9] {

carry = 1;

}

else {

carry = 0;

}

s1.pop[];

s2.pop[];

}

while [!s1.empty[]] {

int sum = carry + s1.top[]->data;

Node* temp = newnode[sum % 10];

s3.push[temp];

if [sum > 9] {

carry = 1;

}

else {

carry = 0;

}

s1.pop[];

}

while [!s2.empty[]] {

int sum = carry + s2.top[]->data;

Node* temp = newnode[sum % 10];

s3.push[temp];

if [sum > 9] {

carry = 1;

}

else {

carry = 0;

}

s2.pop[];

}

// If carry is still present create a new node with

// value 1 and push it to the third stack

if [carry == 1] {

Node* temp = newnode[1];

s3.push[temp];

}

// Link all the elements inside third stack with each

// other

if [!s3.empty[]]

prev = s3.top[];

while [!s3.empty[]] {

Node* temp = s3.top[];

s3.pop[];

if [s3.size[] == 0] {

temp->next = NULL;

}

else {

temp->next = s3.top[];

}

return prev;

}

// utility functions

// Function that displays the List

void Display[Node* head]

{

if [head == NULL] {

return;

}

while [head->next != NULL] {

cout data next;

}

cout data next = NULL;

if [*head_ref == NULL] {

new_node->next = *head_ref;

*head_ref = new_node;

return;

}

Node* last = *head_ref;

while [last->next != NULL && last != NULL] {

last = last->next;

}

last->next = new_node;

return;

}

// Driver Program for above Functions

int main[]

{

// Creating two lists

// first list = 9 -> 5 -> 0

// second List = 6 -> 7

Node* first = NULL;

Node* second = NULL;

Node* sum = NULL;

push[&first, 9];

push[&first, 5];

push[&first, 0];

push[&second, 6];

push[&second, 7];

cout 7 Sum List : 1 -> 0 -> 1 -> 7

Another Approach with time complexity O[N]:

The given approach works as following steps:

First, we calculate the sizes of both the linked lists, size1 and size2, respectively.
Then we traverse the bigger linked list, if any, and decrement till the size of both become the same.
Now we traverse both linked lists till the end.
Now the backtracking occurs while performing addition.
Finally, the head node is returned to the linked list containing the answer.

C++

#include

using namespace std;

struct Node {

int data;

struct Node* next;

};

// recursive function

Node* addition[Node* temp1, Node* temp2, int size1,

int size2]

{

// creating a new Node

Node* newNode

= [struct Node*]malloc[sizeof[struct Node]];

// base case

if [temp1->next == NULL && temp2->next == NULL] {

// addition of current nodes which is the last nodes

// of both linked lists

newNode->data = [temp1->data + temp2->data];

// set this current node's link null

newNode->next = NULL;

// return the current node

return newNode;

}

// creating a node that contains sum of previously added

// number

Node* returnedNode

= [struct Node*]malloc[sizeof[struct Node]];

// if sizes are same then we move in both linked list

if [size2 == size1] {

// recursively call the function

// move ahead in both linked list

returnedNode = addition[temp1->next, temp2->next,

size1 - 1, size2 - 1];

// add the current nodes and append the carry

newNode->data = [temp1->data + temp2->data]

+ [[returnedNode->data] / 10];

}

// or else we just move in big linked list

else {

// recursively call the function

// move ahead in big linked list

returnedNode = addition[temp1, temp2->next, size1,

size2 - 1];

// add the current node and carry

newNode->data

= [temp2->data] + [[returnedNode->data] / 10];

}

// this node contains previously added numbers

// so we need to set only rightmost digit of it

returnedNode->data = [returnedNode->data] % 10;

// set the returned node to the current node

newNode->next = returnedNode;

// return the current node

return newNode;

}

// Function to add two numbers represented by nexted list.

struct Node* addTwoLists[struct Node* head1,

struct Node* head2]

{

struct Node *temp1, *temp2, *ans = NULL;

temp1 = head1;

temp2 = head2;

int size1 = 0, size2 = 0;

// calculating the size of first linked list

while [temp1 != NULL] {

temp1 = temp1->next;

size1++;

}

// calculating the size of second linked list

while [temp2 != NULL] {

temp2 = temp2->next;

size2++;

}

Node* returnedNode

= [struct Node*]malloc[sizeof[struct Node]];

// traverse the bigger linked list

if [size2 > size1] {

returnedNode = addition[head1, head2, size1, size2];

}

else {

returnedNode = addition[head2, head1, size2, size1];

}

// creating new node if head node is >10

if [returnedNode->data >= 10] {

ans = [struct Node*]malloc[sizeof[struct Node]];

ans->data = [returnedNode->data] / 10;

returnedNode->data = returnedNode->data % 10;

ans->next = returnedNode;

}

else

ans = returnedNode;

// return the head node of linked list that contains

// answer

return ans;

}

void Display[Node* head]

{

if [head == NULL] {

return;

}

while [head->next != NULL] {

cout data next;

}

cout data data = d;

new_node->next = NULL;

if [*head_ref == NULL] {

new_node->next = *head_ref;

*head_ref = new_node;

return;

}

Node* last = *head_ref;

while [last->next != NULL && last != NULL] {

last = last->next;

}

last->next = new_node;

return;

}

// Driver Program for above Functions

int main[]

{

// Creating two lists

Node* first = NULL;

Node* second = NULL;

Node* sum = NULL;

push[&first, 5];

push[&first, 6];

push[&first, 3];

push[&second, 8];

push[&second, 4];

push[&second, 2];

cout 4 -> 2 Sum List : 1 -> 4 -> 0 -> 5

Related Article: Add two numbers represented by linked lists | Set 2

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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Practice Tags :

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Linked List

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1. Using Hashing

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