使用地图使用链表添加两个多项式

给定由链表表示的两个多项式数。编写一个函数来计算它们的代数和。

例子：

Input:
1st number = 5x^2 + 4x^1 + 2x^0
2nd number = 5x^1 + 5x^0
Output: 5x^2 + 9x^1 + 7x^0

方法：该实现使用地图数据结构，以便将所有具有相同幂值的系数相加并保存在地图内的键值对中。

下面是上述方法的实现：

// C++ program for addition of two polynomials
// represented as linked lists
#include 
using namespace std;
  
// Structure of Node used
struct Node {
  
    // Coefficient of the polynomial
    int coeff;
  
    // Power of the polynomial
    int pow;
    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;
    }
}
  
// Display Linked list
void show(struct Node* node)
{
    if (node == NULL)
        return;
    while (node->next != NULL) {
        printf("%dx^%d", node->coeff, node->pow);
        node = node->next;
        if (node->next != NULL)
            printf(" + ");
    }
}
  
/* Function to print the required sum of a 
polynomial p1 and p2 as specified in output */
void addPolynomial(Node* p1, Node* p2)
{
    map mp;
    Node* temp1 = p1;
    Node* temp2 = p2;
    while (temp1 != NULL) {
  
        // Add coefficients of same power value
        // map contains (powervalue, coeff) pair
        mp[temp1->pow] += temp1->coeff;
        temp1 = temp1->next;
    }
    while (temp2 != NULL) {
        mp[temp2->pow] += temp2->coeff;
        temp2 = temp2->next;
    }
  
    // Started from the end because higher power should
    // come first and there should be "+" symbol in between
    // so iterate up to second element and print last out of
    // the loop.
  
    for (auto it = (mp.rbegin()); it != prev(mp.rend()); ++it)
        cout << it->second << "x^" << it->first << " + ";
    cout << mp.begin()->second << "x^" << mp.begin()->first;
}
  
// Driver function
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));
  
    printf("\nAdded polynomial: ");
  
    // Function add two polynomial numbers
    addPolynomial(poly1, poly2);
  
    return 0;
}

输出：

1st Number: 5x^2 + 4x^1 + 2x^0
2nd Number: 5x^1 + 5x^0
Added polynomial: 5x^2 + 9x^1 + 7x^0

时间复杂度： O((m + n)log(m+n)) 其中 m 和 n 分别是第一个和第二个列表中的节点数，我们使用地图来添加系数额外的 log(m+n) 因子是添加。

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