链表的建立、插入和删除（三）

文章作者 100test 发表时间 2007:03:10 17:46:25
来源 100Test.Com百考试题网

3. 实例[例7 - 7 ]
创建包含学号、姓名节点的单链表。其节点数任意个，表以学号为序，低学号的在前，高学号的在后，以输入姓名为空作结束。在此链表中，要求删除一个给定姓名的节点，并插入一个给定学号和姓名的节点。
# include "stdlib.h"
# include "malloc. h"
struct node /*节点的数据结构* /
{
int num.
char str[20].
struct node *next.
} .
/ * * * * * * * * * * * * * * * * * * * * * * * * * * * * /
main( )
{
/ *函数声明* /
struct node *creat().
struct node *insert().
struct node *delet().
void print( ).
struct node *head.
char str[20].
int n.
head=NULL. /*做空表* /
head=creat (head). / *调用函数创建以head 为头的链表* /
p r i n t ( h e a d ) ；/ *调用函数输出节点* /
printf("\n input inserted num,name:\n").
gets(str). /*输入学号* /
n=atoi (str).
gets(str). /*输入姓名* /
head=insert (head, str, n). 将/*节点插入链表*/
print (head)； / *调用函数输出节点*/
printf("\n input 0deleted name:\n").
gets(str). /*输入被删姓名* /
head=delet(head,str). /调*用函数删除节点*/
print (head). /*调用函数输出节点* /
r e t u r n .
}
/ * * * * * * * * * * * * * * * * * * * * * * /
/ * * * 创建链表* * * * * * * * * * * * /
struct node *creat(struct node *head)
{
char temp[30].
struct node *pl,*p2.
pl=p2=(struct node*) malloc(sizeof(struct node)).
printf ("input num, name: \n.")
printf("exit:double times Enter!\n").
g e t s ( t e m p ) .
gets (p1->str).
pl->num=atoi (temp).
p l - > n e x t = N U L L .
while (strlen (pl->str)>0
{
if (head==NULL) head=pl；
else p2->next=p1.
P 2 = p l ；
pl=(struct node *)malloc(sizeof(struct node)).
printf ("input num, name: \n").
printf("exit:double times Enter!\n").
g e t s ( t e m p ) .
gets(pl ->str).
p1->num=atoi (temp).
P 1 - > n e x t = N U L L .
}
return head.
}
/ * * * * * * * * * * * * * * * * * * * * /
/ * * * * * * * * * * 插入节点* * * * * * * * * * /
struct node *insert (head, pstr,n).
struct node *head.
char *pstr.
int n.
{
struct node *pl,*p2,*p3.
p1=(struct node*)malloc(sizeof(struct node)).
strcpy (p1->str, pstr).
p 1 - > n u m = n .
p 2 = h e a d .
i f ( h e a d = = N U L L )
{
h e a d = p l . p l - > n e x t = N U L L .
}
e l s e
{
while (n>p2->num&.&.p2->next!=NULL)
{
p 3 = P 2
p 2 = p 2 - > n e x t .
}
if (n<=p2->num)
if (head==p2)
{
h e a d = p l .
p l - > n e x t = p 2 .
}
else
{
p 3 - > n e x t = p l .
p l - > n e x t = p 2 .
}
else
{
p 2 - > n e x t = p l .
p l - > n e x t = N U L L .
}
}
r e t u r n ( h e a d ) .
}