实验11-2-5 链表拼接 (20 分)

1. 题目摘自

https://pintia.cn/problem-sets/13/problems/605

2. 题目内容

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：

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。

输入样例：

1 3 5 7 -1
2 4 6 -1

输出样例：

1 2 3 4 5 6 7

3. 源码参考

#include <iostream>
#include <stdlib.h>

using namespace std;

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

struct ListNode *createlist(); 
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, *t;
  int n;

  h = t = NULL;
  cin >> n;
  while(n != -1)
  {
    p = (struct ListNode*)malloc(sizeof(struct ListNode));
    
    p->data = n;
    p->next = NULL;
    
    if(h == NULL)
    {
      h = p;
    }
    else
    {
      t->next = p;
    }

    t = p;
    
    cin >> n;
  }
  
  return h;
}

struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2)
{
  struct ListNode *p1,*p2,*h,*t,*p;
  int a[100];
  int n;
  int i,j,te;
  
  if((list1==NULL)&&(list2==NULL))
  {
    return NULL;
  }
  
  n=0;
  p1=list1;
  while(p1)
  {
    a[n++]=p1->data;
    p1=p1->next;
  }
  
  p2=list2;
  while(p2)
  {
    a[n++]=p2->data;
    p2=p2->next;
  }
  
  for(i=0;i<n;i++)
  {
    for(j=i+1;j<n;j++)
    {
      if(a[i]>a[j])
      {
        te=a[i];
        a[i]=a[j];
        a[j]=te;
      }
    }
  }
  
  h=t=NULL;
  for(i=0;i<n;i++)
  {
    p=(struct ListNode*)malloc(sizeof(struct ListNode));
    
    p->data=a[i];
    p->next=NULL;
    
    if(h==NULL)
    {
      h=p;
    }
    else
    {
      t->next=p;
    }
    
    t=p;
  }

  return h;
}