分别给定两个单链表形式的多项式 P1和P2 ，任务是以单链表的形式打印商和余数表达式，通过将多项式P1除以P2 得到。
注意：假设多项式表示为x 的较高幂到 x的较低幂(即 0)。
例子：
Input: P1 = 5 -> 4 -> 2, P2 = 5 -> 5
Output:
Quotient = 1 -> 0.2
Remainder = 3
Input: P1 = 3 -> 5 -> 2, P2 = 2 -> 1
Output:
Quotient = 1.5 -> 1.75
Remainder = 0.25
处理方法：按照以下步骤解决问题：
- 创建两个单向链表quotient和剩余部分，其中每个节点将包含 x 的幂系数和指向下一个节点的指针。
- 当余数的次数小于除数的次数时，请执行以下操作：
- 通过在功率因数和商店减去分红的长期领先的功率。
- 将被除数的首项系数除以除数并存储在变量系数 中。
- 从步骤 1和步骤 2 中形成的项创建一个新节点N ，并在商列表中插入 N。
- 将 N 乘以除数，然后从得到的结果中减去被除数。
- 完成上述步骤后，打印商和余数列表。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Node structure containing power and
// coefficient of variable
struct Node {
float coeff;
int pow;
struct Node* next;
};
// Function to create new node
void create_node(float x, int y,
struct Node** temp)
{
struct Node *r, *z;
z = *temp;
// If temp is NULL
if (z == NULL) {
r = (struct Node*)malloc(
sizeof(struct Node));
// Update coefficient and
// power in the LL z
r->coeff = x;
r->pow = y;
*temp = r;
r->next = (struct Node*)malloc(
sizeof(struct Node));
r = r->next;
r->next = NULL;
}
// Otherwise
else {
r->coeff = x;
r->pow = y;
r->next = (struct Node*)malloc(
sizeof(struct Node));
r = r->next;
r->next = NULL;
}
}
// Function to create a LL that stores
// the value of the quotient while
// performing polynomial division
void store_quotient(float mul_c, int diff,
struct Node* quo)
{
// Till quo is non-empty
while (quo->next != NULL) {
quo = quo->next;
}
// Update powers and coefficient
quo->pow = diff;
quo->coeff = mul_c;
quo->next = (struct Node*)malloc(
sizeof(struct Node));
quo = quo->next;
quo->next = NULL;
}
// Function to create a new polynomial
// whenever subtraction is performed
// in polynomial division
void formNewPoly(int diff, float mul_c,
struct Node* poly)
{
// Till poly is not empty
while (poly->next != NULL) {
poly->pow += diff;
poly->coeff *= mul_c;
poly = poly->next;
}
}
// Function to copy one polynomial
// into another linkedlist
void copyList(struct Node* r,
struct Node** copy)
{
// Copy the values of r in the
// polynomial copy
while (r != NULL) {
struct Node* z
= (struct Node*)malloc(
sizeof(struct Node));
// Store coeffcient and power
z->coeff = r->coeff;
z->pow = r->pow;
z->next = NULL;
struct Node* dis = *copy;
if (dis == NULL) {
*copy = z;
}
else {
while (dis->next != NULL) {
dis = dis->next;
}
dis->next = z;
}
r = r->next;
}
}
// Function to subtract two polynomial
void polySub(struct Node* poly1,
struct Node* poly2,
struct Node* poly)
{
// Compute until poly1 and poly2 is empty
while (poly1->next && poly2->next) {
// If power of 1st polynomial
// > 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;
poly->next
= (struct Node*)malloc(
sizeof(struct Node));
poly = poly->next;
poly->next = NULL;
}
// If power of 2nd polynomial >
// 1st then store 2nd as it is
// and move its pointer
else if (poly1->pow < poly2->pow) {
poly->pow = poly2->pow;
poly->coeff = -1 * poly2->coeff;
poly2 = poly2->next;
poly->next
= (struct Node*)malloc(
sizeof(struct Node));
poly = poly->next;
poly->next = NULL;
}
// If power of both polynomial
// is same then subtract their
// coefficients
else {
if ((poly1->coeff
- poly2->coeff)
!= 0) {
poly->pow = poly1->pow;
poly->coeff = (poly1->coeff
- poly2->coeff);
poly->next = (struct Node*)malloc(
sizeof(struct Node));
poly = poly->next;
poly->next = NULL;
}
// Update the pointers
// poly1 and poly2
poly1 = poly1->next;
poly2 = poly2->next;
}
}
// Add the remaining value of polynomials
while (poly1->next || poly2->next) {
// If poly1 exists
if (poly1->next) {
poly->pow = poly1->pow;
poly->coeff = poly1->coeff;
poly1 = poly1->next;
}
// If poly2 exists
if (poly2->next) {
poly->pow = poly2->pow;
poly->coeff = -1 * poly2->coeff;
poly2 = poly2->next;
}
// Add the new node to poly
poly->next
= (struct Node*)malloc(
sizeof(struct Node));
poly = poly->next;
poly->next = NULL;
}
}
// Function to display linked list
void show(struct Node* node)
{
int count = 0;
while (node->next != NULL
&& node->coeff != 0) {
// If count is non-zero, then
// print the postitive value
if (count == 0)
cout << node->coeff;
// Otherwise
else
cout << abs(node->coeff);
count++;
// Print polynomial power
if (node->pow != 0)
cout << "x^" << node->pow;
node = node->next;
if (node->next != NULL)
// If coeff of next term
// > 0 then next sign will
// be positive else negative
if (node->coeff > 0)
cout << " + ";
else
cout << " - ";
}
cout << "\n";
}
// Function to divide two polynomials
void divide_poly(struct Node* poly1,
struct Node* poly2)
{
// Initialize Remainder and Quotient
struct Node *rem = NULL, *quo = NULL;
quo = (struct Node*)malloc(
sizeof(struct Node));
quo->next = NULL;
struct Node *q = NULL, *r = NULL;
// Copy poly1, i.e., dividend to q
copyList(poly1, &q);
// Copy poly, i.e., diivisor to r
copyList(poly2, &r);
// Perform polynomial subtraction till
// highest power of q > highest power of divisor
while (q != NULL
&& (q->pow >= poly2->pow)) {
// difference of power
int diff = q->pow - poly2->pow;
float mul_c = (q->coeff
/ poly2->coeff);
// Stores the quotient node
store_quotient(mul_c, diff,
quo);
struct Node* q2 = NULL;
// Copy one LL in another LL
copyList(r, &q2);
// formNewPoly forms next value
// of q after performing the
// polynomial subtraction
formNewPoly(diff, mul_c, q2);
struct Node* store = NULL;
store = (struct Node*)malloc(
sizeof(struct Node));
// Perform polynomial subtraction
polySub(q, q2, store);
// Now change value of q to the
// subtracted value i.e., store
q = store;
free(q2);
}
// Print the quotient
cout << "Quotient: ";
show(quo);
// Print the remainder
cout << "Remainder: ";
rem = q;
show(rem);
}
// Driver Code
int main()
{
struct Node* poly1 = NULL;
struct Node *poly2 = NULL, *poly = NULL;
// Create 1st Polynomial (Dividend):
// 5x^2 + 4x^1 + 2
create_node(5.0, 2, &poly1);
create_node(4.0, 1, &poly1);
create_node(2.0, 0, &poly1);
// Create 2nd Polynomial (Divisor):
// 5x^1 + 5
create_node(5.0, 1, &poly2);
create_node(5.0, 0, &poly2);
// Function Call
divide_poly(poly1, poly2);
return 0;
} 输出：
Quotient: 1x^1 - 0.2
Remainder: 3
时间复杂度： O(M + N)
辅助空间： O(M + N)
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