题解 | #牛牛的单链表求和#

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

typedef struct list{
    int data;
    struct list *next;
}list,*linklist;
//求和函数
int  get_Sum_List(list *head)
{
    list *point=head;
    int sum=0;
    while(point != NULL)
    {
        sum += point->data;
        point = point->next;
    }
    return sum;
}

#if 1
int main()
{
    int i,n;    
    scanf("%d",&n);
    int arr[n];    
    for(i=0;i<n;i++)
    {
        scanf("%d",&arr[i]); 
    }
    //链表的创建
    list *headList = NULL;
    list *arrList = NULL;
    list *head = NULL;

    head = headList;  //防止 headList 不为空链表的时候，而导致下面循环链表无头 
    //尾插法 
    for(i=0;i<n;i++)
    { 
        arrList = (list *)malloc(sizeof(list));
        arrList->data = arr[i]; 
        arrList->next = NULL; //将headList放在arrList->next,即arrList的数据是放在headList的前面的 
        //headList=head;
        if(headList == NULL)
        {
            headList = arrList;  //
            head = headList;
        }
        else
        {
            while(headList->next != NULL)
            {
                headList = headList->next ;
            }
            headList->next = arrList;
        } 
    }
    //ptink(head);
    printf("%d",get_Sum_List(head));
    
    return 0;    
} 
#endif