题解 | #牛牛的双链表求和#

#include<stdio.h>
#include<stdlib.h>

int N;
int val1,val2;

struct ListNode
{
    int data;
    struct list_node *next;  
};

typedef struct ListNode gut;

gut *creat_list_node(int c_data)//创建结点
{
    gut *New_Node=NULL;
    New_Node=(gut*)malloc(sizeof(gut));
    memset(New_Node,0,sizeof(gut));
    New_Node->data=c_data;
    New_Node->next=NULL;
    
    return New_Node;
}

gut *tail_node(gut *HeadNode,int t_data)//尾插
{
    gut *new_node=creat_list_node(t_data);
    gut *insert_node=HeadNode;
    
    while(insert_node->next!=NULL)
    {
        insert_node=insert_node->next;
    }
    insert_node->next=new_node;
    
    return insert_node;
}

gut *add_node(gut *HeadNode_1,gut *HeadNode_2)//加和
    {
     gut *add_1=HeadNode_1->next;//头结点是0，去除头结点
     gut *add_2=HeadNode_2->next;
    
    while(add_1 !=NULL)//遍历至尾结点
    {
        add_1->data=add_1->data+add_2->data;//加和
        add_1=add_1->next;//下一结点
        add_2=add_2->next;
    }
    
    return add_1;//返回加和后的链表

    }

gut *print_node(gut *HeadNode)
{
    gut *pri_node=HeadNode->next;
    while(pri_node!=NULL)
    {
        printf("%d ",pri_node->data); //打印      
        pri_node=pri_node->next;//下一结点
    }
    printf("\n");
    
    return pri_node->data;//不要忘了返回值
}


int main()
{
    gut *HeadNode1=creat_list_node(N);//创建两个头结点（相当于开辟两个存储空间）
    gut *HeadNode2=creat_list_node(N);
    while(scanf("%d\n",&N)!=EOF)
    {
     for(int i=0;i<N;i++)//链表a
     {
         scanf("%d ",&val1);       
         tail_node(HeadNode1, val1);
     }
         for(int i=0;i<N;i++)//链表b
     {
         scanf("%d ",&val2);
         tail_node(HeadNode2, val2);
     }
        
       add_node(HeadNode1,HeadNode2) ;//加和
       print_node(HeadNode1);//打印
        
    }   
    free(HeadNode1);//释放内存
    free(HeadNode2);
    return 0;
}