题解 | #牛牛的双链表求和#

多写你就明白了
#include<stdio.h>
#include<stdlib.h>
typedef struct Doublelist{
int value;
struct Doublelist*next;
struct Doublelist*prev;
}Doublelist;
Doublelist*creatNode(int value)
{
	Doublelist*newnode = (Doublelist*)malloc(sizeof(Doublelist));
    newnode->value= value;
 newnode->next = NULL;
    return newnode;
}
Doublelist*creatlist(int*array,int n)
{
	if(n==0)
	return NULL;
	Doublelist*head= creatNode(array[0]);
	Doublelist*current = head;
    for(int i = 1;i<n;i++)
    {
    	current->next= creatNode(array[i]);
    	current = current->next;
    }
    return head;
}
int*summry2(Doublelist*head1,Doublelist*head2,int n)
{
	Doublelist*current1 = head1;
	Doublelist*current2 = head2;
	int sum; 
	int*array3=(int*)malloc(n*sizeof(int));
	int i = 0;
	while(current1!=NULL)
	{
		sum = (current1->value)+(current2->value);
		current1 = current1->next;
		current2 = current2->next;
		array3[i] = sum;
		i++;
	}
	return array3;
}
int main()
{
    int n;
    scanf("%d",&n);
    int*array1= (int*)malloc(n*sizeof(int));
    int*array2= (int*)malloc(n*sizeof(int));
    for(int i = 0;i<n;i++)
    {
    	scanf("%d",&array1[i]);
    }
     for(int i = 0;i<n;i++)
    {
    	scanf("%d",&array2[i]);
    }
    Doublelist*head1 = creatlist(array1,n);
      Doublelist*head2= creatlist(array2,n);
	  int*array3= summry2(head1,head2,n);
      for(int i = 0;i<n;i++)
      {
      	 printf("%d ",array3[i]);
      }
      while(head1!=NULL)
      {
      	Doublelist*temp = head1;
      	head1 = head1->next;
      	free(temp);
      }
       while(head2!=NULL)
      {
      	Doublelist*temp = head2;
      	head2 = head2->next;
      	free(temp);
      }
      free(array1);
       free(array2);
	return 0;
}