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 = 5x^{2}+ 4x^{1}+ 2x^{0}2nd number = -5x^{1}- 5x^{0}Output: 5x^{2}-1x^{1}-3x^{0}Input: 1st number = 5x^{3}+ 4x^{2}+ 2x^{0}2nd number = 5x^1 - 5x^0 Output: 5x^{3}+ 4x^{2}+ 5x^{1}- 3x^{0}

## 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 then 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 then 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` `(` `"\n2nd Number: "` `);` ` ` `show(poly2);` ` ` `poly = (` `struct` `Node*)` `malloc` `(` `sizeof` `(` `struct` `Node));` ` ` `// Function add two polynomial numbers` ` ` `polyadd(poly1, poly2, poly);` ` ` `// Display resultant List` ` ` `printf` `(` `"\nAdded polynomial: "` `);` ` ` `show(poly);` ` ` `return` `0;` `}` |

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**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.**Related Article: **Add two polynomial numbers using Arrays

This article is contributed by **Akash Gupta**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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