靜態鍊錶相對動態鍊錶，（基於你指標和結構體，記憶體有清晰的認識的情況下）比較簡單，下面是小生的練習**，提供給大家。
#include #include struct test
;void prinlink(struct test *head)
}int main()
; struct test t2=;
struct test t8=;
struct test t3=;
struct test t4=;
t1.next = &t2;
t2.next = &t8;
t8.next = &t3;
t3.next = &t4;
prinlink(&t1);
system("pause");
return 0;
}
#include #include struct test
;void prinlink(struct test *head)
putchar('\n');
}int chazhao(struct test *head) //查詢節點個數
return cnt;
}int chashu(struct test *head,int data) //查詢有沒有這個點
p = p->next;
}return 0;
}int main()
; struct test t2=;
struct test t3=;
struct test t4=;
struct test t5=;
t1.next = &t2;
t2.next = &t3;
t3.next = &t4;
t4.next = &t5;
prinlink(&t1);
int ret;
ret = chazhao(&t1);
printf("ret = %d",ret);
putchar('\n');
int pon;
pon = chashu(&t1,3);
if(pon == 1)else
putchar('\n');
pon = chashu(&t1,9);
if(pon == 1)else
system("pause");
return 0;
}
#include #include struct test
;void prinlink(struct test *head)
putchar('\n');
}int houm(struct test *head,int data,struct test *new)
p = p->next;
}return 0;
}int main()
; struct test t2 = ;
struct test t3 = ;
struct test t4 = ;
struct test t5 = ;
struct test t6 = ;
t1.next = &t2;
t2.next = &t3;
t3.next = &t4;
t4.next = &t5;
prinlink(&t1);
putchar('\n');
houm(&t1,3,&t6);
prinlink(&t1);
system("pause");
return 0;
}
#include #include struct test
;struct test *qianm(struct test *head,int data,struct test *new)
while(p->next != null)
p = p->next;
}return 0;
}void prinlink(struct test *head)
}int main()
; struct test t2 = ;
struct test t3 = ;
struct test t4 = ;
struct test t5 = ;
t1.next = &t2;
t2.next = &t3;
t3.next = &t4;
t4.next = &t5;
head = &t1;
struct test new = ;
head = qianm(head,5,&new);
prinlink(head);
putchar('\n');
struct test new2 = ;
head = qianm(head,999,&new2);
prinlink(head);
system("pause");
return 0;
}
#include #include struct test
;void prinlink(struct test *head)
}struct test *shanchu(struct test *head,int data)
while(p->next != null)
p = p->next;
}return 0;
}int main()
; struct test t2 = ;
struct test t3 = ;
struct test t4 = ;
struct test t5 = ;
t1.next = &t2;
t2.next = &t3;
t3.next = &t4;
t4.next = &t5;
head = &t1;
prinlink(head);
putchar('\n');
head = shanchu(head,3);
prinlink(head);
putchar('\n');
system("pause");
return 0;
}
**裡面有一些小演算法，看不懂可以多看幾遍o(￣▽￣)ｄ 。
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