链表的逆转

╰+攻爆jí腚メ 2022-01-10 00:29 172阅读 0赞

链表的逆转：巧妙之处在于递归的使用，自我炫耀一番，嘿嘿。  
  
\#include <stdio.h>  
\#include <stdlib.h>  
  
\#define LEN sizeof(struct student)  
\#define FORMANT() \\  
printf("==================================\\n");\\  
printf("num eng math avg(auto)\\n");\\  
printf("==================================\\n")  
  
struct student  
\{  
int num;  
float english;  
float math;  
float avg;  
struct student \*next;  
\};  
  
int n = 0;  
  
struct student \*creat();  
void prt(struct student \*head);  
struct student \*exchange(struct student \*head);  
  
int main(int argc, char \*argv\[\])  
\{  
struct student stu, \*head;  
int pos,position;  
  
printf("创建链表：\\n");  
head = creat();  
prt(head);  
printf("\\n逆转链表：\\n");  
head = exchange(head);  
prt(head);  
  
system("PAUSE");  
return 0;  
\}  
  
struct student \*creat()  
\{  
struct student \*p1, \*p2, \*head = NULL;  
  
p1 = p2 = (struct student \*)malloc(LEN);  
printf("请输入学生的信息：\\n");  
FORMANT();  
scanf("%d%f%f", &p1->num,&p1->english,&p1->math);  
p1->avg = (p1->english + p1->math)/2;  
  
while(p1->num != 0)  
\{  
n++;  
if(n == 1)  
\{  
head = p1;  
\}  
else  
\{  
p2->next = p1;  
\}  
p2 = p1;  
p1 = (struct student \*)malloc(LEN);  
scanf("%d%f%f", &p1->num,&p1->english,&p1->math);  
p1->avg = (p1->english + p1->math)/2;  
\}  
p1->next = NULL;  
return head;  
\}  
  
void prt(struct student \*head)  
\{  
struct student \*p1;  
p1 = head;  
printf("\\n显示链表:\\nnow there are %d datas.\\n",n);  
FORMANT();  
while(p1 != NULL)  
\{  
printf("%-8d%-8.2f%-9.2f%-8.2f\\n", p1->num, p1->english, p1->math, p1->avg);  
p1=p1->next;  
\}  
\}  
  
struct student \*exchange(struct student \*head)  
\{  
struct student \*p1, \*p2;  
  
p1 = head;  
if(p1->next == NULL)  
\{  
head = p1;  
\}  
else  
\{  
p2 = p1->next;  
head = exchange(p2);  
p2->next = p1;  
p1->next = NULL;  
\}  
return head;  
\}

转载于:https://blog.51cto.com/qq164587043/78007