题解 | #牛牛的双链表求和#

#include <stdio.h>

typedef struct node{
    int data;
    struct node *next;
}node;
int sum_node(node *head);
void printf_node(node *head);
void add_node(node *head,int data);
node* sum_double_node(node* head_1,node* head_2);
node *cteat_node();
int main() {
    int n,e,count;
    node *head_1=cteat_node();
    node *head_2=cteat_node();
    scanf("%d ",&n);
    
    count =0;
    while (scanf("%d ", &e) != EOF) {   
        count++;
        add_node(head_1,e);
        if(count == n)break;
   }
        count =0;
       while (scanf("%d ", &e) != EOF) { 
  
        add_node(head_2,e);
       
   }
   node *head_3=sum_double_node(head_1,head_2);
    printf_node(head_3);
   //printf_node(head_1);
    return 0;
}

//创建头链表  
//返回值为链表表头
//链表头默认data=0；
//链表尾->next = NULL
node *cteat_node(){
    node *head=(node *)malloc(sizeof(node));
    head->next=NULL;
    return head;
}

//尾插添加
// head为链表头  data为数据
void add_node(node *head,int data){
    node *pT=NULL;
    node *new=(node *)malloc(sizeof(node));
    new->data=data;
    for(pT=head;pT->next!=NULL;pT=pT->next){};//表尾->next为NULL
    pT->next=new;//链接
}

//打印链表
//head为表头
void printf_node(node *head){
    
    for(node *pT=head;pT!=NULL;pT=pT->next){
        if(pT!=head)printf("%d ",pT->data);
    }
}

//函数功能：求和
//函数参数：head 表头 
//函数返回值：int sum  
int sum_node(node *head){
    int sum=0;//求和值sum
    for(node *pT= head;pT != NULL; pT= pT->next ){
        sum+=pT->data;
    }//循环到pT=NULL sum累加
    return sum;
}
//function : 链表2连接到链表1后
//parame ： 链表1表头head_1  链表2表头 head_2 

node* sum_double_node(node* head_1,node* head_2){
        node *head_3=cteat_node();
        int count1=0;
        int count2=0;
        for(node *pT_1= head_1; pT_1 != NULL;pT_1=pT_1->next){ 
            count2=0;
             count1++;
             for(node *pT_2= head_2; pT_2 != NULL;pT_2=pT_2->next)  {count2++; 
                if(count1 == count2  && count1 !=1) add_node(head_3,pT_1->data+pT_2->data);
             }
    }
    return head_3;
}