实验11-2-5 链表拼接（20 分）

本题要求实现一个合并两个有序链表的简单函数。链表结点定义如下：

struct ListNode {
    int data;
    struct ListNode *next;
};

函数接口定义：

1	struct ListNode mergelists(struct ListNode list1, struct ListNode *list2);

其中list1和list2是用户传入的两个按data升序链接的链表的头指针；函数mergelists将两个链表合并成一个按data升序链接的链表，并返回结果链表的头指针。

裁判测试程序样例：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};

struct ListNode *createlist(); /*裁判实现，细节不表*/
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);
void printlist( struct ListNode *head )
{
     struct ListNode *p = head;
     while (p) {
           printf("%d ", p->data);
           p = p->next;
     }
     printf("\n");
}

int main()
{
    struct ListNode  *list1, *list2;

    list1 = createlist();
    list2 = createlist();
    list1 = mergelists(list1, list2);
    printlist(list1);
	
    return 0;
}

/* 你的代码将被嵌在这里 */

输入样例：

1 2	1 3 5 7 -1 2 4 6 -1

输出样例：

1	1 2 3 4 5 6 7

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};

struct ListNode *createlist(); /*裁判实现，细节不表*/
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);
void printlist( struct ListNode *head )
{
     struct ListNode *p = head;
     while (p) {
           printf("%d ", p->data);
           p = p->next;
     }
     printf("\n");
}

int main()
{
    struct ListNode  *list1, *list2;

    list1 = createlist();
    list2 = createlist();
    list1 = mergelists(list1, list2);
    printlist(list1);
	
    return 0;
}
struct ListNode *createlist()
{
    struct ListNode *p,*h,*q;   
    p=q=(struct ListNode*)malloc(sizeof(struct ListNode));
    h=(struct ListNode*)malloc(sizeof(struct ListNode));
    int data;   //存放数据
    h=p;    //h为头节点
    scanf("%d",&data);
    while(data!=-1)
    {
        p->data=data;    //保存数据
        q->next=p;   //p指向下一个节点
        q=p;
        p=(struct ListNode*)malloc(sizeof(struct ListNode));
        scanf("%d",&data);
    }
    q->next=NULL;  //尾节点Next指向空
    free(p);
    return h;
}
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2)
{
    int num = 0;
    int temp[100];
    struct ListNode  *p = list1;
    while(p != NULL)
    {
        temp[num] = p->data;
        num++;
        p = p->next;
    }
    p = list2;
    while(p != NULL)
    {
        temp[num] = p->data;
        num++;
        p = p->next;
    }
    int i,j;
    for(i = 0; i < num; i++)
        for(j = i + 1; j < num; j++)
        {
            if(temp[i] > temp[j])
            {
                int t;
                t = temp[i];
                temp[i] = temp[j];
                temp[j] = t;
            }
        }
      struct ListNode  *newlist = NULL;
      struct ListNode  *endlist = NULL;
      struct ListNode  *q;
      for(i = 0; i < num; i++)
      {
          q = (struct ListNode  *)malloc(sizeof(struct ListNode));
          q->data = temp[i];
          if(newlist == NULL)
          {
              newlist = q;
              newlist->next = NULL;
          }
            if(endlist != NULL)
         {
            endlist->next = q;
         }
         endlist = q;
         endlist->next = NULL;
      }
      return newlist;
}