请教一道题。这题是建立两个链表，再把它们连接起来，就是在第一条链表的结尾连接上第二条链表，下面是我写的程序，为什么打印出的数据少了两条链表的最后一个数据？
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#include<stdio.h>
#include<stdlib.h>

#define M 4
#define N 3
#define LEN sizeof(struct ListData)

struct ListData
{
int num;
struct ListData *next;
};

typedef struct ListData LIST;

int n;

struct ListData * __CreatList(int count)
{
n = 0;

struct ListData *head;
struct ListData *p1, *p2;

p1 = p2 = (struct ListData *)malloc(LEN);
if (p1 == NULL)
{
printf("Can't get values.\n");
exit (1);
}

head = NULL;

printf("Input the data:\n");
scanf("%d", &p1->num);

while (count)
{
n++;

if (n == 1)
head = p1;
else
p2->next = p1;

p2 = p1;

p1 = (struct ListData *)malloc(LEN);
if (NULL == p1)
{
exit (1);
}

printf("Input the data:\n");
scanf("%d", &p1->num);

count--;
}

p2->next = NULL;

return (head);
}

struct ListData *__LinkList(LIST *L1, LIST *L2)
{
struct ListData *P0, *head;

head = L1->next;

while (L1)
{
P0 = L1;
L1 = L1->next;
}

P0->next = L2->next;

free(L2);

return (head);
}

void __PrintList(struct ListData *P)
{
struct ListData *L;

L = P;

printf("Now the list_data is:\n");
if (P != NULL)
{
while (L != NULL)
{
printf("%d ", L->num);
L = L->next;
}
}
}

int main(void)
{
struct ListData *L1, *L2, *head;

L1 = __CreatList(M);
L2 = __CreatList(N);

head = __LinkList(L1, L2);

__PrintList(head);

system("Pause");
return 0;
}

这里我改成这样了，但是不知道为什么，在__CreatList这个函数中，传递的参数是链表的长度，但在输入数的时候却需要多输入一个数，例如这例子两条链表共输入9个数，但我想输入的只是7个数，这是为什么？

struct ListData *__LinkList(LIST *L1, LIST *L2)
{
struct ListData *P0, *head;

head = L1;
while (L1)
{
P0 = L1;
L1 = L1->next;
}

P0->next = L2;
return (head);
}