SDUT 2119-数据结构实验之链表四：有序链表的归并

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 5
1 23 26 45 66 99
14 21 28 50 100

Example Output

1 14 21 23 26 28 45 50 66 99 100

#include<stdio.h>#include<stdlib.h>typedef struct node{    int data;    struct node *next;}ST;ST *creat(int n){    ST *head, *tail, *p;    head = (ST *)malloc(sizeof(ST));    head->next = NULL;    tail = head;    while(n--)    {        p = (ST *)malloc(sizeof(ST));        scanf("%d", &p->data);        p->next = NULL;        tail->next = p;        tail = p;    }    return head;}void merge(ST *head, ST *head1){    ST *q, *p, *tail;    q = head->next;    p = head1->next;    free(head1);    tail = head;    while(q && p)    {        if(q->data > p->data)        {            tail->next = p;            tail = p;            p = p->next;        }        else        {            tail->next = q;            tail = q;            q = q->next;        }    }    if(q)    tail->next = q;    else tail->next = p;}void input(ST *head){    ST *tail;    for(tail = head->next; tail != NULL; tail = tail->next)    {        printf("%d", tail->data);        if(tail->next != NULL) printf(" ");        else printf("\n");    }}int main(){    ST *head, *head1;    int m, n;    scanf("%d %d", &m, &n);    head = creat(m);    head1 = creat(n);    merge(head, head1);    input(head);    return 0;}