链表的逆序输出与两个有序链表的合并

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

struct node{
    int data;
    struct node *next;
};

int main()
{
    struct node *head , *p , *q , *t;
    int i , n , a;
    void reverse(struct node *t);
    scanf("%d",&n);
    head = NULL;
    for(i = 0;i < n;i++)
    {
        scanf("%d",&a);
        p = (struct node *)malloc(sizeof(struct node));
        p->data = a;
        p->next = NULL;
        if(head == NULL)
        {
            head = p;
        }
        else
        {
            q->next = p;
        }
        q = p;
    }
    t = head;
    reverse(t);
    printf("\n");
    return 0;
}
//递归输出
void reverse(struct node *l)
{
    if(l != NULL)
    {
        reverse(l->next);
        printf("%d ",l->data);
    }
}

确实好久没碰C语言了，感觉好生疏，本来数据结构的基础不是很好，现在可以趁机会补补数据结构了。

下面是将两个有序链表合并成一个有序链表 :

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

struct node{
    int data;
    struct node *next;
};

void input(struct node *p,int n);
void and_list(struct node *list1,struct node *list2);
void output(struct node *l);

int main()
{
    int n;
    struct node *p , *q;
    p = (struct node *)malloc(sizeof(struct node));
    q = (struct node *)malloc(sizeof(struct node));
    p->next = NULL;
    q->next = NULL;
    printf("please input the number of the number in the first linklist : \n");
    scanf("%d",&n);
    input(p,n);
    printf("please input the number of the number in the second linklist : \n");
    scanf("%d",&n);
    input(q,n);
    and_list(p,q);
    output(p);
    printf("\n");
}

void input(struct node *p,int n)
{
    struct node *temp , *end;
    int i;
    end = p;
    for(i = 0;i < n;i++)
    {
        temp = (struct node *)malloc(sizeof(struct node));
        scanf("%d",&temp->data);
        end->next = temp;
        temp->next = NULL;
        end = temp;
    }
}

void and_list(struct node *list1,struct node *list2)
{
    struct node *p , *q , *temp;
    p = list1;
    q = list2->next;
    while(p->next && q)
    {
        if(p->next->data > q->data)
        {
            temp = q->next;
            q->next = p->next;
            p->next = q;
            q = temp;
        }
        else
        {
            p = p->next;
        }
    }
    if(q)
    {
        p->next = q;
    }
}

void output(struct node *list)
{
    while(list->next)
    {
        printf("%d ",list->next->data);
        list = list->next;
    }
}