使用链表添加两个多项式的 C++ 程序

给定两个由链表表示的多项式数。编写一个添加这些列表的函数意味着添加具有相同变量幂的系数。
例子：

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 
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;
}

输出：

1st Number: 5x^2+4x^1+2x^0
2nd Number: -5x^1-5x^0
Added polynomial: 5x^2-1x^1-3x^0

时间复杂度： O(m + n)，其中 m 和 n 分别是第一个和第二个列表中的节点数。

递归方法：

算法：

1. 如果两个数字都为空，则返回
2. 否则，如果比较幂，如果相同，则添加系数并在两个数字的下一个元素上递归调用 addPolynomials。
3. 否则，如果第一个数字的幂更大，则打印第一个数字的当前元素，并在第一个数字的下一个元素和第二个数字的当前元素上递归调用 addPolynomial。
4. 否则打印第二个数字的当前元素，并在第一个数字的当前元素和第二个数字的下一个元素上递归调用 addPolynomial。

C++

//Program to add two polynomials represented in linkedlist using recursion
#include
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

输出：

Linked List
 5x^2 4x^1
Linked List
 6x^2 4x^1
Addition:
 11x^2  8x^1 

时间复杂度： O(m + n)，其中 m 和 n 分别是第一个和第二个列表中的节点数。

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