两个链表相加 结果不对

[fdg个sggggg]

就是两个多项式相加，幂数高的在前相等的相加，一正一负的相减，最后按幂数的高低输出结果正确的结果是
6
7
7
4
5
1
7
0
这个链表


#include <iostream>
#include <stdlib.h>

using namespace std;
struct itm
{
int a;
int m;
itm *next;
};
itm *insert1()
{
itm *p,*first,*pp;
p=NULL;
first=p;
int a1[4]={6,-4,2,7};
int m1[4]={5,3,1,0};
for(int i=0;i <=3;i++)
{
p=new itm;
if(first==NULL) first = p;
else pp->next = p;
p->a=a1[i];
p->m=m1[i];
p->next = NULL;
pp = p;
}

return first;
}
itm *insert2()
{
itm *q,*second,*pq;
q=NULL;
second=q;
int a2[3]={7,8,3};
int m2[3]={4,3,1};
for(int j=0;j <=2;j++)
{
q=new itm;
if(second==NULL) second = q;
else pq->next = q;
q->a=a2[j];
q->m=m2[j];
q->next = NULL;
pq = q;
}

return second;
}
itm *slect(itm *&m,itm *&n)
{
itm *p,*q,*phead,*qhead;
phead=m;
qhead=n;
p=phead;
q=qhead;
while(q!=NULL)
{
if(p->m>q->m)
{
phead=p;
p=p->next;
}
if(p->m <q->m)
{
if(phead->m <qhead->m)
{
qhead->next=phead;
phead=qhead;
qhead=p->next;
delete qhead;
}
else
{
phead->next=q;
q=p;
q=q->next;
delete q;
}
}
if(p->m==q->m)
{
if(p->a+q->a==0)
{
phead->next=p->next;
delete p;
q=q->next;
delete q;
}
else
{
p->a=(p->a+q->a);
q=q->next;
delete q;
}
}
return phead;
}
}
void print(itm *phead)
{
itm *r=phead;
while(r!=NULL)
{
cout < <r->a < <endl;
cout < <r->m < <endl;
r=r->next;
}
}
int main(int argc, char *argv[])
{
itm *p,*q,*r;
p=insert1();
q=insert2();
r=slect(p,q);
print(r);

system("PAUSE");
return 0;
}

[fdg个sggggg]

谢了

[OenAuth.Core]

来晚了

[wangqiang8848]

我写了一个和你的要求一样的程序，大约写了5K，要的话给我发邮件 wng

[wangqiang8848]

[e-mail]

[liubuweiright]

代码长,不太会

顶

[coverallwangp]

#include <iostream> 
#include <stdlib.h> 

using namespace std; 
struct itm 
{ 
    int a; 
    int m; 
    itm *next; 
}; 
itm *insert1() 
{ 
    itm *p,*first,*pp; 
    p=NULL; 
    first=p;  
    int a1[4]={6,-4,2,7}; 
    int m1[4]={5,3,1,0};      
    for(int i=0;i <=3;i++) 
    {  
        p=new itm; 
        if(first==NULL) first = p; 
        else pp->next = p; 
        p->a=a1[i]; 
        p->m=m1[i]; 
        p->next = NULL; 
        pp = p; 
    }  

    return first; 
} 
itm *insert2() 
{  
    itm *q,*second,*pq; 
    q=NULL; 
    second=q; 
    int a2[3]={7,8,3}; 
    int m2[3]={4,3,1}; 
    for(int j=0;j <=2;j++) 
    {  
        q=new itm; 
        if(second==NULL) second = q; 
        else pq->next = q; 
        q->a=a2[j]; 
        q->m=m2[j]; 
        q->next = NULL; 
        pq = q; 
    }  

    return second; 
} 
itm *slect(itm *&m,itm *&n) 
{ 
    itm *p,*q,*phead,*qhead; 
    phead=NULL; 
    qhead=phead; 
    p=m; 
    q=n; 
    while(q!=NULL&&p!=NULL) 
    { 
        if(p->m > q->m) 
        { 
			if(qhead==NULL)
			{
				qhead=p;
                phead=qhead;				
			}
			else
               qhead->next=p;
            p=p->next;
			
        } 
        if(p->m < q->m) 
        { 
                if(qhead==NULL)
			    {
					qhead=p;
					phead=qhead;				
			    }
			    else
                   qhead->next=q;                
                q=q->next;                 
				qhead = qhead->next;
           
        } 
        if(p->m == q->m) 
        { 
            if(p->a + q->a==0) 
            {                 
                p=p->next;				
                q=q->next; 
                
            } 
            else 
            { 
                p->a=(p->a+q->a);
                if(qhead==NULL)
			    {
					qhead=p;
					phead=qhead;				
				}
			    else                
				    qhead->next=p;
				qhead=qhead->next;
				q=q->next; 
				p=p->next;
               			
            }        
        } 
            
    } 
	if (p!=NULL)
	{
		qhead->next=p;
	}
	if (q!=NULL)
	{
		qhead->next=q;
	}

	return phead;  
	  
}      
void print(itm *phead) 
{ 
    itm *r=phead; 
    while(r!=NULL) 
    { 
        cout <<r->a <<endl; 
        cout <<r->m <<endl; 
        r=r->next;    
    }                  
}          
int main(int argc, char *argv[]) 
{ 
    itm *p,*q,*r; 
    p=insert1(); 
    q=insert2(); 
    r=slect(p,q); 
    print(r); 
    
    system("PAUSE"); 
    return 0; 
} 

你的程序理解起来比较费劲，所以自己改写了一下。希望对你有所帮助。

[coverallwangp]

6
5
7
4
4
3
5
1
7
0
结果如上