两个有序链表序列的合并 (20 分)

已知两个非降序链表序列S1与S2，设计函数构造出S1与S2合并后的新的非降序链表S3。

输入格式:
输入分两行，分别在每行给出由若干个正整数构成的非降序序列，用−1表示序列的结尾（−1不属于这个序列）。数字用空格间隔。

输出格式:
在一行中输出合并后新的非降序链表，数字间用空格分开，结尾不能有多余空格；若新链表为空，输出NULL。

输入样例:

1 3 5 -1
2 4 6 8 10 -1

输出样例:

1 2 3 4 5 6 8 10

题目思路：常规建立链表思路。当然，这道题还能使用vector容器解决，更加快捷。后续有时间再写一篇博客发出来。

#include<iostream>
using namespace std;

typedef struct node {
  int data;
  struct node* next;
}List;

List* creat();
List* sort(List* L1, List* L2);
void print(List* L);
void Delete(List* L);
int main()
{
  List* L1, * L2, *L3;

  L1 = creat();
  L2 = creat();
  L3 = sort(L1, L2);
  print(L3);

  Delete(L3);
  return 0;
}
List* creat()
{
  List* head, * p, * rear;
  head = new List;
  head->next = NULL;
  rear = head;


  p = new List;
  p->next = NULL;
  cin >> p->data;
  while (p->data != -1)
  {
    rear->next = p;
    rear = p;


    p = new List;
    p->next = NULL;
    cin >> p->data;
  }

  return head;
}
void print(List* L)
{
  List* p;
  p = L->next;
  if (p)
  {
    cout << p->data;
    p = p->next;
    while (p)
    {
      cout << " " << p->data;
      p = p->next;
    }
  }
  else
    cout << "NULL";
  
  cout << endl;
}
void Delete(List* L)
{
  List* p, *q;
  p = L;
  while (p)
  {
    q = p->next;
    delete p;
    p = q;
  }

  
}
List* sort(List* L1, List* L2)
{
  List* L3, * pa, * pb, * pc;
  L3 = new List;
  pa = L1->next;
  pb = L2->next;
  pc = L3;
  while (pa && pb)
  {
    if (pa->data < pb->data)
    {
      pc->next = pa;
      pc = pa;
      pa = pa->next;
    }
    else
    {
      pc->next = pb;
      pc = pb;
      pb = pb->next;
    }
  }
  pc->next = pa ? pa : pb;
  
  return L3;
}