C++ Program For Adding Two Polynomials Using Linked List
Last Updated :
23 Jul, 2025
Given two polynomial numbers represented by a linked list. Write a function that add these lists means add the coefficients who have same variable powers.
Example:
Input:
1st number = 5x2 + 4x1 + 2x0
2nd number = -5x1 - 5x0
Output:
5x2-1x1-3x0
Input:
1st number = 5x3 + 4x2 + 2x0
2nd number = 5x^1 - 5x^0
Output:
5x3 + 4x2 + 5x1 - 3x0

CPP
// C++ program for addition of two
// polynomials using Linked Lists
#include <bits/stdc++.h>
using namespace std;
// Node structure containing power
// and coefficient of variable
struct Node
{
int coeff;
int pow;
struct Node* next;
};
// Function to create new node
void create_node(int x, int y,
struct Node** temp)
{
struct Node *r, *z;
z = *temp;
if (z == NULL)
{
r = (struct Node*)malloc(sizeof(struct Node));
r->coeff = x;
r->pow = y;
*temp = r;
r->next = (struct Node*)malloc(sizeof(struct Node));
r = r->next;
r->next = NULL;
}
else
{
r->coeff = x;
r->pow = y;
r->next = (struct Node*)malloc(sizeof(struct Node));
r = r->next;
r->next = NULL;
}
}
// Function Adding two polynomial
// numbers
void polyadd(struct Node* poly1,
struct Node* poly2,
struct Node* poly)
{
while (poly1->next &&
poly2->next)
{
// If power of 1st polynomial is greater
// than 2nd, then store 1st as it is and
// move its pointer
if (poly1->pow > poly2->pow)
{
poly->pow = poly1->pow;
poly->coeff = poly1->coeff;
poly1 = poly1->next;
}
// If power of 2nd polynomial is greater
// than 1st, then store 2nd as it is and
// move its pointer
else if (poly1->pow < poly2->pow)
{
poly->pow = poly2->pow;
poly->coeff = poly2->coeff;
poly2 = poly2->next;
}
// If power of both polynomial numbers
// is same then add their coefficients
else
{
poly->pow = poly1->pow;
poly->coeff = (poly1->coeff +
poly2->coeff);
poly1 = poly1->next;
poly2 = poly2->next;
}
// Dynamically create new node
poly->next =
(struct Node*)malloc(sizeof(struct Node));
poly = poly->next;
poly->next = NULL;
}
while (poly1->next || poly2->next)
{
if (poly1->next)
{
poly->pow = poly1->pow;
poly->coeff = poly1->coeff;
poly1 = poly1->next;
}
if (poly2->next)
{
poly->pow = poly2->pow;
poly->coeff = poly2->coeff;
poly2 = poly2->next;
}
poly->next =
(struct Node*)malloc(sizeof(struct Node));
poly = poly->next;
poly->next = NULL;
}
}
// Display Linked list
void show(struct Node* node)
{
while (node->next != NULL)
{
printf("%dx^%d", node->coeff,
node->pow);
node = node->next;
if (node->coeff >= 0)
{
if (node->next != NULL)
printf("+");
}
}
}
// Driver code
int main()
{
struct Node *poly1 = NULL,
*poly2 = NULL,
*poly = NULL;
// Create first list of 5x^2 +
// 4x^1 + 2x^0
create_node(5, 2, &poly1);
create_node(4, 1, &poly1);
create_node(2, 0, &poly1);
// Create second list of -5x^1 -
// 5x^0
create_node(-5, 1, &poly2);
create_node(-5, 0, &poly2);
printf("1st Number: ");
show(poly1);
printf("2nd Number: ");
show(poly2);
poly = (struct Node*)malloc(sizeof(struct Node));
// Function add two polynomial
// numbers
polyadd(poly1, poly2, poly);
// Display resultant List
printf("Added polynomial: ");
show(poly);
return 0;
}
Output1st Number: 5x^22nd Number: -5x^1Added polynomial: 5x^2-5x^1
Output:
1st Number: 5x^2+4x^1+2x^0
2nd Number: -5x^1-5x^0
Added polynomial: 5x^2-1x^1-3x^0
Time Complexity: O(m + n) where m and n are number of nodes in first and second lists respectively.
Auxiliary Space: O(max(m,n) as it is using extra space for resultant linked list "poly"
Recursive Method :
Algorithm:
- If both the numbers are null then return
- else if compare the power, if same then add the coefficients and recursively call addPolynomials on the next elements of both the numbers.
- else if the power of the first number is greater then print the current element of the first number and recursively call addPolynomial on the next element of the first number and current element of the second number.
- else print the current element of the second number and recursively call addPolynomial on the current element of first number and next element of second number.
C++
//Program to add two polynomials represented in linkedlist using recursion
#include<iostream>
using namespace std;
class Node{
public:
int coeff,power;
Node *next;
Node(int coeff, int power)
{
this->coeff = coeff;
this->power = power;
this->next = NULL;
}
};
void addPolynomials(Node *head1,
Node *head2)
{
if(head1==NULL &&
head2==NULL)
return;
else if(head1->power ==
head2->power)
{
cout << " " << head1->coeff + head2->coeff <<
"x^" << head1->power << " ";
addPolynomials(head1->next,head2->next);
}
else if(head1->power > head2->power)
{
cout << " " << head1->coeff <<
"x^" << head1->power << " ";
addPolynomials(head1->next, head2);
}
else
{
cout << " " << head2->coeff <<
"x^" << head2->power << " ";
addPolynomials(head1, head2->next);
}
}
void insert(Node *head,
int coeff,
int power)
{
Node *new_node =
new Node(coeff,power);
while(head->next != NULL)
{
head = head->next;
}
head->next = new_node;
}
void printList(Node *head)
{
cout << "Linked List" << endl;
while(head != NULL)
{
cout << " " << head->coeff <<
"x" << "^" << head->power;
head = head->next;
}
}
// Driver code
int main()
{
Node *head=new Node(5, 2);
insert(head, 4, 1);
Node *head2 = new Node(6, 2);
insert(head2, 4, 1);
printList(head);
cout << endl;
printList(head2);
cout << endl <<
"Addition:" << endl;
addPolynomials(head,head2);
return 0;
}
//This code is contributed by Akshita Patel
OutputLinked List
5x^2 4x^1
Linked List
6x^2 4x^1
Addition:
11x^2 8x^1
Time Complexity: O(m + n) where m and n are number of nodes in first and second lists respectively.
Auxiliary Space: O(max(m,n))
Please refer complete article on Adding two polynomials using Linked List for more details!
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