数据结构实验之链表四：有序链表的归并_数据结构实验之链表四:有序链表的归并-优快云博客

该博客介绍了如何根据输入的两个有序整数序列构建两个有序单链表，并将它们合并成一个大的有序链表。实验要求不使用数组，重点在于链表的操作和排序算法的应用。给出的示例展示了具体输入和输出格式。

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think:
1顺序建立链表+链表的顺序输出+链表的有序(顺序)归并;链表的元素插入

sdut原题链接

数据结构实验之链表四：有序链表的归并
Time Limit: 1000MS Memory Limit: 65536KB

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

Hint
不得使用数组！

Author

以下为accepted代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct node
{
    int Data;
    struct node *next;
};
struct node * Build(int n);//顺序建立链表
struct node * ans(struct node *head1, struct node *head2);//链表的有序(顺序)归并
void Pri(struct node *head);//链表的顺序输出
int main()
{
    int n, m;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        struct node *head1, *head2, *head;
        head1 = (struct node *)malloc(sizeof(struct node));
        head1->next = NULL;
        head2 = (struct node *)malloc(sizeof(struct node));
        head2->next = NULL;
        head = (struct node *)malloc(sizeof(struct node));
        head->next = NULL;
        head1 = Build(n);
        //Pri(head1);
        head2 = Build(m);
        //Pri(head2);
        head = ans(head1, head2);
        Pri(head);
    }
    return 0;
}
void Pri(struct node *head)//链表的顺序输出
{
    struct node *p;
    p = head->next;
    while(p != NULL)
    {
        printf("%d%c", p->Data, p->next == NULL? '\n': ' ');
        p = p->next;
    }
}
struct node * Build(int n)//顺序建立链表
{
    struct node *head, *tail;
    head = (struct node *)malloc(sizeof(struct node));
    head->next = NULL;
    tail = head;
    struct node *p;
    for(int i = 0; i < n; i++)
    {
        p = (struct node *)malloc(sizeof(struct node));
        scanf("%d", &p->Data);
        p->next = tail->next;
        tail->next = p;
        tail = p;
    }
    return head;
}
struct node * ans(struct node *head1, struct node *head2)//链表的有序(顺序)归并
{
    struct node *head, *tail;
    head = (struct node *)malloc(sizeof(struct node));
    head->next = NULL;
    tail = head;
    struct node *p, *q, *a, *b;
    p = head1->next;
    a = head2->next;
    q = p, b = a;
    while(p != NULL && a != NULL)
    {
        if(p->Data < a->Data)
        {
            q = p->next;
            p->next = tail->next;
            tail->next = p;
            tail = p;
            p = q;

            if(p == NULL)
                break;
        }
        else
        {
            b = a->next;
            a->next = tail->next;
            tail->next = a;
            tail = a;
            a = b;

            if(a == NULL)
                break;
        }
    }
    if(p == NULL)
        tail->next = a;
    else if(a == NULL)
        tail->next = p;
    free(head1);
    free(head2);
    return head;
}


/***************************************************
User name: 
Result: Accepted
Take time: 0ms
Take Memory: 108KB
Submit time: 2017-03-10 17:07:53
****************************************************/