sdutacm-数据结构实验之链表六：有序链表的建立

来源：互联网发布：js pdf 插件编辑：程序博客网时间：2024/06/06 03:22

数据结构实验之链表六：有序链表的建立

TimeLimit: 1000MS Memory Limit: 65536KB

SubmitStatistic

ProblemDescription

输入N个无序的整数，建立一个有序链表，链表中的结点按照数值非降序排列，输出该有序链表。

Input

第一行输入整数个数N；
第二行输入N个无序的整数。

Output

依次输出有序链表的结点值。

ExampleInput

33 6 22 9 44 5

ExampleOutput

5 6 9 22 33 44

Hint

不得使用数组！

Author

F#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<iostream>#include<algorithm>#include<stack>#include<queue>//#include<dqeue>using namespace std;struct node{  int data;  struct node*next;};struct  node*creat(struct node*head,int n){      head= (struct node*)malloc(sizeof(struct node));      head->next =NULL;      struct node*tail = head;      for(int i=1;i<=n;i++)      {          struct node*p = (struct node*)malloc(sizeof(struct node));          p->next =NULL;         scanf("%d",&p->data);         tail->next = p;         tail = p;      }      return head;}/*struct node*guibing(struct node*head1,struct node*head2){    head1= head1->next;    head2 =head2->next;    struct node*tail,*p,*mail;    tail = (struct node*)malloc(sizeof(struct node));    mail =  tail;    while(head1&&head2)    {        if(head1->data<head2->data)        {           mail->next = head1;           mail = head1;           p = head1;           head1 = head1->next;           p->next =NULL;        }        else        {          mail->next =head2;          mail = head2;          p = head2;          head2 = head2->next;          p->next =NULL;        }        }        if(head1)        {            while(head1)            {                mail->next =head1;                mail = head1;                p = head1;                head1 = head1->next;                p->next = NULL;            }        }        else if(head2)        {          while(head2)          {             mail->next =head2;             mail = head2;             p = head2;             head2 = head2->next;             p->next =NULL;          }    }return tail;}*/struct node*search(struct node*head1,int n){   while(n--)   {      struct node*tail = head1;      int key;      cin>>key;      while(tail->next)      {         if(tail->next->data>key)         {            struct node*p;            p = (struct node*)malloc(sizeof(struct node));            p->next = NULL;            p->data = key;            p->next = tail->next;            tail->next = p;            tail = head1;            break;         }         else tail = tail->next;      }      if(tail->next==NULL)      {           struct node*p = (struct node*)malloc(sizeof(struct node));           p->next = NULL;           p->data = key;           tail->next = p;      }   }   return head1;}void show(struct node*head){     struct node*tail;     tail =head;     int top = 1;     while(tail->next)     {           if(top)top=0;           else printf(" ");           printf("%d",tail->next->data);           tail = tail->next;     }}int main(){        int n,n1=0,n2=0;        scanf("%d",&n);        struct node*head1= (struct node*)malloc(sizeof(struct node));       head1->next =NULL;       head1 = search(head1,n);       show(head1);       cout<<endl;         return 0;        }/***************************************************User name: jk160505徐红博Result: AcceptedTake time: 0msTake Memory: 164KBSubmit time: 2017-01-14 17:07:53****************************************************/

0 0