Problem D: 逆置链式链表(线性表)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 594 Solved: 346
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Description
本题只需要提交填写部分的代码
(线性表)试编写算法将线性表就地逆置,以链式存储结构实现。
代码:
#include <stdio.h>
#include <malloc.h>
struct Num
{
int n;
struct Num *next;
}num;
struct Num *createlist(struct Num *head);
void print(struct Num *head);
void destroy(struct Num *head);
void destroy(struct Num *head)
{
struct Num *p;
while(head!=NULL)
{
p=head->next;
delete(head);
head=p;
}
}
#include <malloc.h>
struct Num
{
int n;
struct Num *next;
}num;
struct Num *createlist(struct Num *head);
void print(struct Num *head);
void destroy(struct Num *head);
void destroy(struct Num *head)
{
struct Num *p;
while(head!=NULL)
{
p=head->next;
delete(head);
head=p;
}
}
int main()
{
struct Num *head=NULL;
head=createlist(head); //建立
print(head);//输出
destroy(head);
return 0;
}
struct Num *createlist(struct Num *head) //头插法建立链表
{
struct Num *p;
p=head=(struct Num*)malloc(sizeof(struct Num));
head=NULL;
p=(struct Num*)malloc(sizeof(struct Num)); //p建立新结点
while(scanf("%d",&p->n)!=EOF) //将新结点插到开头的位置
{
/***************/
添加代码
/*****************/
p=(struct Num*)malloc(sizeof(struct Num));
//p每次建立新结点
}
return head;
}
}
return head;
}
void print(struct Num *head)
{
struct Num *current=head;
while(current!=NULL)
{
printf("%d ",current->n);
current=current->next;
}
}
{
struct Num *current=head;
while(current!=NULL)
{
printf("%d ",current->n);
current=current->next;
}
}
Input
1 2 3 4 5 6 7 8 9
Output
9 8 7 6 5 4 3 2 1
Sample Input
10 23 56 89 11
Sample Output
11 89 56 23 10
p->next=head; head=p;