链表合并（多个函数写法，思路清晰易理解）

解题思路:1.分别建立链表1和2；

              2.连接链表1、2；

              3.连接后的总链表为链表3；

              4.用选择排序法对链表3进行排序。

注意事项:    注意多个头指针及next指针的指向（此处为难点）

参考代码:

#include<stdio.h>

#include<stdlib.h>

typedef struct node

{

        int id;

        int sc;

        struct node *next;

}Node;

Node *input1(int m)

{

    Node *p1,*p2,*head1;                                                                                                                                                                                                                                                                                 

    int i;

    for(i=0;i<m;i++)

    {

        p1=(Node *)malloc(sizeof(Node));

        if(i==0)

        {

            scanf("%d %d",&p1->id,&p1->sc);

            head1=p1;

            p2=p1;

        }

        else

        {

            scanf("%d %d",&p1->id,&p1->sc);

            p2->next=p1;

            p2=p1;

        }

    }

    p2->next=NULL;

    return head1;

}

Node *input2(int n)

{

    Node *p1,*p2,*head2;

    int i;

    for(i=0;i<n;i++)

    {

        p1=(Node *)malloc(sizeof(Node));

        if(i==0)

        {

            scanf("%d %d",&p1->id,&p1->sc);

            head2=p1;

            p2=p1;

        }

        else

        {

            scanf("%d %d",&p1->id,&p1->sc);

            p2->next=p1;

            p2=p1;

        }

    }

    p2->next=NULL;

    return head2;

}

Node *connect(Node *head1,Node *head2)

{

    Node *head3=head1;

    while(head1->next!=NULL)

    {

        head1=head1->next;

    }

    head1->next=head2;

    return head3;  

}

void paixu(Node *head3)         //选择排序法

{

    Node *min,*last;

    int a,b;

    for(min=head3;min;min=min->next)

    {

        for(last=min->next;last;last=last->next)

        {

            if(min->id >last->id )

            {

                a=min->id ;min->id =last->id ;last->id =a;

                b=min->sc ;min->sc =last->sc ;last->sc =b;

            }

        }

    }

}

void print(Node *head3)

{

    while(head3)

    {

        printf("%d %d\n",head3->id,head3->sc);

        head3=head3->next; 

    }

}

int main()

{

    int m,n;

    scanf("%d %d",&m,&n);

    Node *head1=input1(m);

     Node *head2=input2(n);

    Node *head3=connect(head1,head2);

    paixu(head3);

    print(head3);

    return 0;

}