链表多项式乘法

Posted 2023-02-16

【中文标题】链表多项式乘法

【英文标题】：Linked list polynomial multiplication

【发布时间】：2017-03-08 03:55:24

【问题描述】：

我编写了一个程序，以单链表的形式将两个多项式相乘。我无法使用此代码打印输出。

我得到的输出是

1st Number: 5x^2 + 4x^1 + 2x^0
2nd Number: 5x^1 + 5x^0
Multiplied polynomial:

我该如何解决这个问题？

我的代码：

// C++ program for multiplication 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 exp;
    struct node *next;
;
void padd(float, int, node**);          
// 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->exp = y;
        *temp = r;
        r->next = (struct node*)malloc(sizeof(struct node));
        r = r->next;
        r->next = NULL;
    
    else
    
        r->coeff = x;
        r->exp = y;
        r->next = (struct node*)malloc(sizeof(struct node));
        r = r->next;
        r->next = NULL;
    


// Function Multiplying two polynomial numbers
void polymul ( struct node *poly1, struct node *poly2,
                   struct node *poly )

    struct node *y1 ;
    float coeff1;
    int exp1 ;

    y1 = poly2 ;  /* point to the starting of the second linked list */
if ( poly1 == NULL && poly2 == NULL )
        return ;

    /* if one of the list is empty */
if ( poly1 == NULL )
        poly = poly2 ;
    else
    
        if ( poly2 == NULL )
            poly = poly1 ;
        else/* if both linked lists exist */

        
            /* for each term of the first list */
            while ( poly1 != NULL )
            
                /* multiply each term of the second linked list with a 
                    term of the first linked list */
                while ( poly2 != NULL )
                
                    coeff1 = poly1 -> coeff * poly2 -> coeff ;
                    exp1 = poly1 -> exp + poly2 -> exp ;
                    poly2 = poly2 -> next ;

                    /* add the new term to the resultant polynomial */

                    padd ( coeff1, exp1, &poly ) ;
                

                poly2 = y1 ;  /* reposition the pointer to the starting of
                         the second linked list */


                poly1 = poly1 -> next ;  /* go to the next node */

            
        
    


/* adds a term to the polynomial in the descending order of the exponent */
void padd ( float coeff, int exp, struct node **s )

    struct node *r, *temp = *s ;

    /* if list is empty or if the node is to be inserted before the first node */
if ( *s == NULL || exp > ( *s ) -> exp )
    
        *s=r = (struct node*) malloc ( sizeof ( struct node ) ) ;
        (*s) -> coeff = coeff ;
        (*s) -> exp = exp ;
        (*s)-> next = temp ;
    
    else
    
        /* traverse the entire linked list to search the position to insert a new node */
        while ( temp != NULL )
            
                if ( temp -> exp == exp )
                
                    temp -> coeff += coeff ;
                    return ;
                

                if ( temp -> exp > exp && ( temp -> next -> exp < exp ||    temp -> next == NULL ) )
                    
                        r = (struct node* )malloc ( sizeof ( struct node ) ) ;
                        r -> coeff = coeff;
                        r -> exp = exp ;
                        r -> next = temp -> next ;
                        temp -> next = r ;
                        return ;
                    

            temp = temp -> next ;  /* go to next node */

            

        r -> next = NULL ;
        temp -> next = r ;
    


// Display Linked list
void show(struct node *node)

while(node->next != NULL)
    
    printf("%dx^%d", node->coeff, node->exp);
    node = node->next;
    if(node->next != NULL)
        printf(" + ");
    


// Driver program
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 multiply two polynomial numbers
    polymul(poly1, poly2, poly);

    // Display resultant List
    printf("\nMultiplied polynomial: ");
    show(poly);

return 0;

【问题讨论】：

附带说明：声明struct 会自动使struct 名称成为类型名称，并且您不需要重复struct 字，另外，更喜欢new / @987654327 @ 超过malloc / free。 malloc 不调用构造函数，new 不需要强制转换（C 中的 malloc 也不需要）。

【参考方案1】：

你应该在函数polymul之前声明你的函数padd()，或者你可以在顶部写一个函数原型，以便编译器知道函数padd将在polymul之后遇到。

对于第二个错误，您应该将 malloc 函数的输出类型转换为 struct node *，因为 malloc 返回 void * 类型值。

只需将类型转换添加到代码中的 malloc 函数即可。

r = (struct node*)malloc ( sizeof ( struct node ) ) ;

链表中存在一些指针问题。我已经在 ideone 链接中修复了它们。

 #include<bits/stdc++.h>

   using namespace std;

// Node structure containing power and coefficient of variable

   struct node
   
    int coeff;
    int exp;
    struct node *next;
;

// Function to create new node


void create_node(int x, int y, struct node **temp)


    struct node *r, *tmp=NULL ;

    if(*temp == NULL) // If the list is NULL then add node at First position
    
        r =(struct node*)malloc(sizeof(struct node));
        r->coeff = x;
        r->exp = y;
        r->next = NULL;
        *temp = r;
        /*r->next = (struct node*)malloc(sizeof(struct node));
        r = r->next;
        r->next = NULL;*/ //Not needed
    
    else
    
        tmp = *temp;
        while (tmp->next != NULL) 
            tmp = tmp -> next; //Travel to the end of linked list
        
        r = (struct node*)malloc(sizeof(struct node)); 
        r->coeff = x;
        r->exp = y;
        r->next = NULL;
        tmp->next = r; //Add new node to the list

    


// Function Adding two polynomial numbers

 /* adds a term to the polynomial in the descending order of the exponent */

    void padd ( int coeff, int exp, struct node **s )
   
    struct node *r, *temp = *s ;
    struct node *tmp = *s;
    /* if list is empty or if the node is to be inserted before the first node 
    */



    if ( temp == NULL || exp > ( temp ) -> exp )

    
        r = (struct node*)malloc ( sizeof ( struct node ) ) ;
        r -> coeff = coeff ;
        r -> exp = exp ;
        r -> next = temp ;
        temp = r;
        *s = r;
    
    else
    
        /* traverse the entire linked list to search the position to insert 
        a new node */
        while ( temp != NULL )
            
                if ( temp -> exp == exp )
                
                    temp -> coeff += coeff ;
                    return;
                 else if ( temp -> exp > exp && ( temp -> next -> exp < exp ||temp ->next == NULL ) )
                    
                        r = (struct node*)malloc ( sizeof ( struct node ) ) ;
                        r -> coeff = coeff;
                        r -> exp = exp ;
                        r -> next = temp -> next ;
                        temp -> next = r ;
                        return;
                    

                temp = temp -> next ;  /* go to next node */

            

        r -> next = NULL ;
        temp -> next = r ;
    


   struct node * polymul ( struct node *poly1, struct node *poly2, struct node *poly3)
   

    struct node *y1 ;
    int coeff1;
    int exp1 ;

    struct node *poly = poly3;
    y1 = poly2 ;         /* point to the starting of the second linked list */
    if ( poly1 == NULL && poly2 == NULL )
        return NULL;

    /* if one of the list is empty */
    if ( poly1 == NULL ) 
        poly = poly2 ;
     else
    
        if ( poly2 == NULL )
            poly = poly1 ;
        else    /* if both linked lists exist */

        
            /* for each term of the first list */
            while ( poly1 != NULL )
            
                /* multiply each term of the second linked list with a 
                    term of the first linked list */
                while ( poly2 != NULL )
                
                    coeff1 = poly1 -> coeff * poly2 -> coeff ;
                    exp1 = poly1 -> exp + poly2 -> exp ;
                    poly2 = poly2 -> next ;

                    /* add the new term to the resultant polynomial */

                    padd( coeff1, exp1, &poly) ;
                

                poly2 = y1 ;  /* reposition the pointer to the starting of
                                the second linked list */


                poly1 = poly1 -> next ;  /* go to the next node */

            
        
    
    return poly;




// Display Linked list
    void show(struct node *node)
    
        while(node != NULL) // if we use node->next != NULL while will break at the last node skipping it
        
            printf("%dx^%d", node->coeff, node->exp);

            if(node->next != NULL)
                printf(" + ");

            node = node->next;
        
    

// Driver program

    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

    poly = polymul(poly1, poly2, poly);

    // Display resultant List

    printf("\nAdded polynomial: ");
    show(poly);

return 0;

Ideone Link

【讨论】：

它确实解决了我的错误。但是现在，我无法获得两个多项式相乘的所需输出。

@JaiPrakash 您的链接列表中有几个错误。你的ceate_node 函数错了，你的show 函数错了。您需要了解链表并自己调试代码。

我看不出我的 create_node () 函数和 show() 函数是如何出错的。我认为唯一的问题是与此代码混淆的指针。

@JaiPrakash 您创建了 2 个多项式 5x^2 + 4x^1 + 2x^0 和 5x^1 + 5x^0，但显示功能仅打印第一个元素。您的 create node 没有将节点附加到列表中。此外，您的 show 函数不会打印列表的最后一个节点。首先，更正您无法正常工作的代码。

@JaiPrakash 现在检查代码。我已经修复了所有的错误。但是你在指针上犯了很多错误。了解指针，否则您将再次遇到错误。