C 数据结构实验之链表四：有序链表的归并 SDUT

Time Limit: 1000 ms Memory Limit: 65536 KiB

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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。

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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。

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Output

输出合并后的单链表所包含的M+N个有序的整数。

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Sample Input

6 5
1 23 26 45 66 99
14 21 28 50 100

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Sample Output

1 14 21 23 26 28 45 50 66 99 100

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Hint

不得使用数组！

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Source

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先定义两个链表，在定义第三个链表，并对前两个链表从第一个开始比较，那一个小，边连接在第三个链表上；

#include <stdio.h>
#include <stdlib.h>
struct node
{
    int a;
    struct node *next;
};
int main()
{
    int m,n,i,a1;
    struct node *head1,*head2,*head3,*q,*p,*i1;
    scanf("%d%d",&m,&n);
    head1 = (struct node *)malloc(sizeof(struct node));
    head1 -> next = NULL;
    q = head1;
    for(i=0; i<m; i++)
    {
        scanf("%d",&a1);
        p = (struct node *)malloc(sizeof(struct node));
        p -> next = NULL;
        p -> a  = a1;
        q -> next = p;
        q = p;

    }  //定义第一个链表；
    head2 = (struct node *)malloc(sizeof(struct node));
    head2 -> next = NULL;
    q = head2;
    for(i=0; i<n; i++)
    {
        scanf("%d",&a1);
        p = (struct node *)malloc(sizeof(struct node));
        p -> next = NULL;
        p -> a  = a1;
        q -> next = p;
        q = p;

    }   //定义第二个链表；
    head3 = (struct node *)malloc(sizeof(struct node));
    head3 -> next = NULL;
    i1 = head1 -> next;
    q = head2 -> next;
    p = head3;
    free(head1);
    free(head2);
    for(i=0;i<n+m;i++)
    {
      if(i1&&q)
      {
         if(i1 -> a < q -> a)
        {
            p -> next = i1;
            p = i1;
            i1 = i1 -> next;
        }
        else
        {

            p -> next= q;
            p = q;
            q = q -> next;
        }
      }
      else
      {
          if(i1)
            p -> next =i1;
          if(q)
            p -> next = q;
      }
    }
    p = head3 -> next;
     while(p) 
     {
         if(p -> next)
         printf("%d ",p -> a);
         else printf("%d\n",p -> a);
         p = p -> next;

     }
    return 0;
}