将两个升序排列的链表合并后升序排序打印.

╰+攻爆jí腚メ 2022-06-16 11:56 228阅读 0赞

说的再多，我不如放码过来，大家看不懂哪里，可以评论，尴尬忘了注释了。。。

\#include<stdio.h>  
\#include<stdlib.h>  
\#include<string.h>  
\#define LEN sizeof(struct Student)  
struct Student  
\{  
long num;  
float score;  
struct Student \*next;  
\};  
  
  
int n; //全局变量n、用记录链表的结点数.  
  
  
struct Student \*creat(void); //创建链表的函数原型声明.  
  
  
void print(struct Student \*head); //输出链表的函数原型声明.  
  
  
struct Student \*merge(struct Student \*La,struct Student \*Lb); //对两个带有头结点链表的函数声明.  
  
  
int main()  
\{  
printf("Tip : The two LinkLists should be ascending order !\\n\\n");  
struct Student \*p1,\*p2,\*p;  
printf("Create A. ");  
p1=creat();  
print(p1);  
printf("Create B. ");  
p2=creat();  
print(p2);  
p=merge(p1,p2);  
printf("\\nTip: Linklist A and B sorted.\\n");  
print(p);  
  
return 0;  
\}  
  
  
struct Student \*creat(void)  
\{  
struct Student \*head;  
struct Student \*p1,\*p2;  
n=0;  
head=(struct Student \*)malloc(LEN);  
p1=p2=(struct Student \*)malloc(LEN);  
head->next=NULL;  
printf("Please input num and score of students ： \\n");  
scanf("%ld%f",&p1->num,&p1->score);  
while(p1->num!=0) //当输入的X.num!=0 的时候链入链表.  
\{  
n=n+1; //结点数增加.  
if(n==1) head->next=p1;  
else p2->next=p1;  
p2=p1;  
p1=(struct Student \*)malloc(LEN);  
scanf("%ld%f",&p1->num,&p1->score);  
\}  
p2->next=NULL;  
free(p1);  
return(head);  
\}  
  
  
void print(struct Student \*head)  
\{  
struct Student \*p;  
p=head->next;  
printf("Now,These %d records are :\\n",n);  
if(head->next!=NULL)  
\{  
do  
\{  
printf("%-12ld%-12.2f\\n",p->num,p->score);  
p=p->next;  
\}while(p!=NULL);  
\}  
\}  
  
  
struct Student \*merge(struct Student \*La,struct Student \*Lb) //该代码是根据学生的学号来排序的.  
\{  
struct Student \*a,\*b,\*c;  
c=La; //让定义的结构体指针C指向A链表.然后再A表的原有基础之上链入B表的结点.  
a=La->next;  
b=Lb->next;  
while(a!=NULL && b!=NULL)  
\{  
if(a->num <= b->num)  
\{  
c->next=a;  
c=a;  
a=a->next;  
n++;  
\}  
else  
\{  
c->next=b;  
c=b;  
b=b->next;  
n++;  
\}  
\}  
if(a!=NULL) c->next=a;  
else c->next=b;  
  
free(Lb);  
return La;  
\}