C++ Program To Add Two Numbers Represented By Linked Lists- Set 1
Last Updated :
11 Apr, 2023
Given two numbers represented by two lists, write a function that returns the sum list. The sum list is a list representation of the addition of two input numbers.
Example:
Input:
List1: 5->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
Method 1:
Approach: Traverse both lists and One by one pick nodes of both lists and add the values. If the sum is more than 10 then make carry as 1 and reduce sum. If one list has more elements than the other then consider the remaining values of this list as 0. The steps are:
- Traverse the two linked lists from start to end
- Add the two digits each from respective linked lists.
- If one of the lists has reached the end then take 0 as its digit.
- Continue it until both the end of the lists.
- If the sum of two digits is greater than 9 then set carry as 1 and the current digit as sum % 10
Below is the implementation of this approach.
C++
// C++ program to add two numbers
// represented by linked list
#include <bits/stdc++.h>
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 of 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;
}
// A utility function to print a
// linked list
void printList(Node* node)
{
while (node != NULL)
{
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
// Driver code
int main(void)
{
Node* res = NULL;
Node* first = NULL;
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);
cout << "Second List is ";
printList(second);
// Add the two lists and see
// result
res = addTwoLists(first,
second);
cout << "Resultant list is ";
printList(res);
return 0;
}
// This code is contributed by rathbhupendra
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 stack data structure
Approach:
1. Create 3 stacks namely s1,s2,s3.
2. Fill s1 with Nodes of list1 and fill s2 with nodes of list2.
3. 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 .
4. If the sum >9, set carry 1
5. Else set carry 0.
6. Create a Node(say prev) that will contain the head of the sum List.
7. Link all the elements of s3 from top to bottom.
8. return prev.
C++
// C++ program to add two numbers represented
// by Linked Lists using Stack
#include <bits/stdc++.h>
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<Node*> 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 << head->data << " -> ";
head = head->next;
}
cout << head->data << endl;
}
// Function that adds element at
// the end of the Linked List
void push(Node** head_ref, int d)
{
Node* new_node = newnode(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 code
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 << "First List : ";
Display(first);
cout << "Second List : ";
Display(second);
sum = addTwoNumbers(first, second);
cout << "Sum List : ";
Display(sum);
return 0;
}
Output:
First List : 9 -> 5 -> 0
Second List : 6 -> 7
Sum List : 1 -> 0 -> 1 -> 7
Time Complexity: O(n), where n is the length of the longer of the two linked lists.
Space Complexity: O(n), where n is the length of the longer of the two linked lists. We use two stacks to store the nodes of the two linked lists which take up O(n) space in total.
Another Approach with time complexity O(N):
The given approach works as following steps:
- First, we calculate sizes of both the linked lists, size1 and size2, respectively.
- Then we traverse the bigger linked list, if any, and decrement till size of both become same.
- Now we traverse both linked lists till end.
- Now the backtracking occurs while performing addition.
- Finally, the head node is returned of the linked list containing the answer.
C++
// C++ program to implement
// the above approach
#include <iostream>
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 nod
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 << head->data << " -> ";
head = head->next;
}
cout << head->data << endl;
}
// Function that adds element at the
// end of the Linked List
void push(Node** head_ref, int d)
{
Node* new_node =
(struct Node*)malloc(sizeof(struct Node));
new_node->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 code
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 << "First List : ";
Display(first);
cout << "Second List : ";
Display(second);
sum = addTwoLists(first, second);
cout << "Sum List : ";
Display(sum);
return 0;
}
// This code is contributed by Dharmik Parmar
Output:
First List : 5 -> 6 -> 3
Second List : 8 -> 4 -> 2
Sum List : 1 -> 4 -> 0 -> 5
Time complexity: O(n), as we traverse through both linked list and add the digits until we reach the end of the longer list.
Space complexity: O(1), as we are only using a single node to store the sum of the two digits in the current nodes.
Related Article: Add two numbers represented by linked lists | Set 2
Please refer complete article on Add two numbers represented by linked lists | Set 1 for more details!
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