题解 | #牛牛的单向链表#

#include <stdio.h>
typedef struct ListNode* Ptrnode;
struct ListNode{
    int data;
    Ptrnode NEXT;
};
Ptrnode BuildList(int n, int a[]);
void ShowList(Ptrnode P, int n);
int main(void)
{
   int n, i;
   Ptrnode P;
   scanf("%d\n",&n);
   int a[n];
   for(i = 0; i < n; i++)
   {
    scanf("%d",&a[i]);
   }
   P = BuildList(n, a);
   ShowList(P, n);

   return 0;
}

Ptrnode BuildList(int n, int a[])
{
    
    Ptrnode Pp  = (Ptrnode) malloc(sizeof(struct ListNode));
    Pp->data = 0;
    Pp->NEXT = NULL;
    Ptrnode tmp = Pp;
    for(int i = 0; i < n; i++)
    {
        Ptrnode Pm = (Ptrnode) malloc(sizeof(struct ListNode));
        Pm ->data = a[i];
        Pm ->NEXT = NULL;
        Pp ->NEXT  = Pm;
        Pp = Pm;
    }
    return tmp->NEXT;
}

void ShowList(Ptrnode P, int n)
{
    for(int i = 0; i < n; i++)
    {
        printf("%d ",P->data);
        P = P->NEXT;
    }
}