数据结构实验之链表四：有序链表的归并

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Problem Description
分别输入两个有序的整数序列（分别包含M和N个数据），建立两个有序的单链表，将这两个有序单链表合并成为一个大的有序单链表，并依次输出合并后的单链表数据。
Input
第一行输入M与N的值；
第二行依次输入M个有序的整数；
第三行依次输入N个有序的整数。
Output
输出合并后的单链表所包含的M+N个有序的整数。
Example Input

6 51 23 26 45 66 9914 21 28 50 100

Example Output

1 14 21 23 26 28 45 50 66 99 100

Hint
不得使用数组！
Author

#include <stdio.h>#include <stdlib.h>#include <string.h>#include<bits/stdc++.h>#include <iostream>#include <algorithm>#define maxn 10100using namespace std;typedef struct list{    int data;    struct list *next;}sqlist;sqlist *newlist(int n){    sqlist *head;    head=(sqlist *)malloc(sizeof(sqlist));    head->next=NULL;    scanf("%d", &head->data);    n--;    sqlist *mail=head;    while(n--)    {        sqlist *p;        p=(sqlist *)malloc(sizeof(sqlist));        p->next=NULL;        scanf("%d", &p->data);        mail->next=p;        mail=p;    }    return head;}sqlist *add(sqlist *head1, sqlist *head2, sqlist *head3){    sqlist *mail=head3;    while(head1&&head2)    {        if(head1->data<head2->data)        {            mail->next=head1;            mail=mail->next;            sqlist *p;            p=head1;            head1=head1->next;            p->next=NULL;        }        else        {            mail->next=head2;            mail=mail->next;            sqlist *p;            p=head2;            head2=head2->next;            p->next=NULL;        }    }    if(head1)    {        while(head1)        {            mail->next=head1;            mail=mail->next;            sqlist *p=head1;            head1=head1->next;            p->next=NULL;        }    }    else if(head2)    {        while(head2)        {            mail->next=head2;            mail=mail->next;            sqlist *p=head2;            head2=head2->next;            p->next=NULL;        }    }    return head3;}void print(sqlist *head3){    int top=1;    while(head3->next)    {        if(top)top=0;        else printf(" ");        printf("%d", head3->next->data);        head3=head3->next;    }}int main(){    int m, n;    scanf("%d%d", &m, &n);    sqlist *head1, *head2, *head3;    head1=newlist(m);    head2=newlist(n);    head3=(sqlist *)malloc(sizeof(sqlist));    head3->next=NULL;    head3=add(head1, head2, head3);    print(head3);    return 0;}

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