这是关于单链表。我有以下代码。我想现在扩展这些代码，以便可以在某个特定位置添加/删除元素。我不知道如何实施它。
这就是我到目前为止：
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
struct list
{
int amount;
struct element *first;
};
struct element
{
int number;
struct element *next;
int *temp;
};
int main()
{
struct list lk;
struct element *ptr, *temp;
int amount;
int i;
printf("How much elements u want enter?");
scanf("%d", &amount);
ptr = (struct element *) calloc(1, sizeof(struct element));
lk.first = ptr;
lk.amount = 0;
printf("Please enter 1. number :");
scanf("%d", &(ptr->number));
temp = ptr;
for (i = 2; i <= amount; i++)
{
printf("Please enter %d. number", i);
ptr = (struct element *) calloc(1, sizeof(struct element));
lk.amount++;
scanf("%d", &(ptr->number));
ptr->next = NULL;
temp->next = ptr;
temp = ptr;
}
ptr = lk.first;
while (ptr != NULL)
{
printf("%d \n", ptr->number);
ptr = ptr->next;
}
getch();
return 0;
}
我找到了以下方法，但我不知道如何为我的程序调整它：
void insertInList (list* L, element* position, element* new)
{
if (position == 0)
{
new->next = L->first;
L->first= new;
L->amount++;
}
else
{
new->next = position->next;
position->next = new;
L->amount++;
}
}
我在用户输入后对此进行了测试：
struct list lk;
struct element *ptr, *temp, number1, number2;
int amount;
int i;
printf("Which element u want add:");
scanf("%d", number1.number);
printf("On which position u want add the element?:");
scanf("%d", number2.number);
initList(&lk);
insertInList(&lk, &zahl2, &zahl1);
我在行＆gt;之后得到一个AccessViolentException scanf("%d", number1.number);
答案 0 :(得分：0)
将元素插入列表的功能没问题，但我有一些建议：
struct list* L代替list* L（或者您添加了一个typedef，例如typedef struct list LIST;，允许您编写LIST * L而不是struct list* L）。我写了以下函数：
void initList(struct list* L)
{
if (L)
{
L->amount = 0;
L->first = NULL;
}
}
int insertInList(struct list* L, struct element* position, struct element* newElem)
{
int result = 0;
struct element *iterator;
if (L)
{
/* Check if newElem is already within list and if so don't add it again! */
for (iterator = L->first; iterator != NULL; iterator = iterator->next)
{
if (iterator == newElem)
{
break;
}
}
if (iterator != newElem) /* newElem not within list, yet? */
{
if (position != 0)
{
newElem->next = position->next;
position->next = newElem;
}
else
{
newElem->next = L->first;
L->first= newElem;
}
L->amount++;
result = 1;
}
}
return (result);
}
int deleteFromList(struct list* L, struct element* elem)
{
int result = 0;
struct element *iterator;
if ((L) && (L->amount > 0)) /* list with elements in it? */
{
if (L->first != elem) /* elem is not the first element? */
{
/* iterator all items to find entry preceeding elem */
for (iterator = L->first; iterator != NULL; iterator = iterator->next)
{
if (iterator->next == elem)
{
iterator->next = elem->next; /* set next element to elemen after elem */
result = 1;
break;
}
}
}
else
{
L->first = elem->next; /* set new head of list */
result = 1;
}
if (result == 1) /* item deleted? */
{
L->amount--;
elem->next = NULL; /* ensure next pointer of elem does not point into list! */
}
}
return (result);
}
int deleteIndexFromList(struct list* L, int iElement) /* iElement is zero based: 0=first element, 1=second element, ... */
{
int result = 0;
struct element *iterator;
if ((L) && (L->amount >= iElement))
{
/* iterator all items to find entry preceeding elem */
for (iterator = L->first; iterator != NULL; iterator = iterator->next)
{
if (iElement == 0)
{
result = deleteFromList(L, iterator);
break;
}
iElement--;
}
}
return (result);
}
您可以使用调试器和以下测试程序测试代码：
int main(void)
{
struct list L;
struct element A, B, C, D;
A.number = 5;
B.number = 10;
C.number = 15;
D.number = 20;
initList(&L);
insertInList(&L, NULL, &A);
insertInList(&L, &A, &B);
insertInList(&L, &B, &C);
insertInList(&L, &C, &D);
/* now your list is 5 -> 10 -> 15 -> 20 */
deleteFromList(&L, &A);
/* now your list is 10 -> 15 -> 20 */
deleteFromList(&L, &C);
/* now your list is 10 -> 20 */
return (0);
}
答案 1 :(得分：0)
要在“条目N”之后添加条目，您只需要搜索“条目N”。那你就做new_entry->next = current_entry->next; current_entry->next = new_entry;。
要在“条目N”之前添加条目，并删除“条目N”，您需要在搜索时跟踪上一条目的地址;这样当您找到“条目N”时，您仍然知道上一个条目的地址。
在这种情况下，删除“条目N”主要变为previous_entry->next = current_entry->next;;并在“条目N”之前插入新条目变为new_entry->next = previous_entry->next; previous_entry->next = new_entry;。但是，对于这两者，对于“N == 0”，前一个条目可能不存在。在这种情况下，您必须修改列表的头部而不是条目。
例如（删除，其中something是结构类型的名称）：
int deleteEntry(int n, something **headPtr) {
something *previous = NULL;
something *current = *headPtr;
while( (current != NULL) && (n > 0) ) {
previous = current;
current = current->next;
n--;
}
if(current == NULL) {
return ERROR; // End of list reached before N found
}
if(previous == NULL) {
*headPtr = current->next;
} else {
previous->next = current->next;
}
free(current);
return OK;
}
注意：完全可以在n == 0循环之前处理while情况。我这样编写它就可以很容易地修改搜索条件并将其转换为删除满足其他条件的第一个条目的东西。
答案 2 :(得分：0)
让我们从单独链接的非循环列表的2个特殊情况开始。第一种是添加数据的通用方法，只是继续将节点添加到列表的末尾。一个函数可能看起来像：
/* insert node at end of list */
void insert_end (list *l, int n)
{
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
ptr->next = NULL;
if (l->cnt == 0)
{
l->first = ptr;
l->cnt++;
return;
}
lnode *iter = l->first; /* pointer to iterate list */
while (iter->next) iter = iter->next;
iter->next = ptr;
l->cnt++;
}
上面你只是为下一个元素分配存储空间（我保留了节点）。您只需检查金额（重命名为cnt）是否为0。如果是，请添加为第一个节点。如果没有，则创建指向列表的指针以用作迭代器并迭代列表指针，直到node->next为NULL并添加新的节点在最后。
（注意：如果插入效率是关键，使用双链接循环列表，则不需要迭代，只需在list->prev位置添加一个节点即可，即使在拥有数亿个节点的列表中，也会使加法盲目快速）
下一个变体是希望在列表的开头或开头添加新节点。在这里，您只需ptr->next = l->first然后l->first = ptr：
/* insert node at beginning of list */
void insert_start (list *l, int n)
{
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
if (l->cnt == 0)
ptr->next = NULL;
else
ptr->next = l->first;
l->first = ptr;
l->cnt++;
}
如何在列表中的给定位置插入节点。您需要验证位置(0 <= pos <= lk->cnt)（或者您可以设置任何大于lk->cnt的值等于lk->cnt）。您已经了解了如何使用列表指针迭代节点以到达列表的末尾以迭代iter = lter->next，直到到达最后一个节点。到达nth节点并没有什么不同。要在给定位置插入，您将获得位置，因此只需迭代pos次即可到达插入点：
/* insert node at position */
void insert_pos (list *l, int n, int pos)
{
/* validate position */
if (pos < 0 || pos > l->cnt) {
fprintf (stderr, "%s() error: invalid position.\n", __func__);
return;
}
/* if empty or pos 0, insert_start */
if (l->cnt == 0 || pos == 0) {
insert_start (l, n);
return;
}
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
ptr->next = NULL;
lnode *iter = l->first; /* pointer to iterate list */
while (--pos)
iter = iter->next;
if (iter->next)
ptr->next = iter->next;
iter->next = ptr;
l->cnt++;
}
下一个变体是按数字排序顺序在列表中的任何位置添加节点。如果新的ptr->number = 6;已经有5和7，那么请在ptr和5之间插入新的7。 注意：下面的这个函数还处理放置第一个节点和小于列表中第一个节点的节点，以及将节点放在列表的末尾。它基本上可以找到给定新节点的位置。如果您的目标是按排序顺序插入节点，或者您可以使用它来填充特殊情况，则可以将此作为唯一的输入例程。
/* insert node at end of list */
void insert_ordered (list *l, int n)
{
/* if first node of n < first->number */
if (l->cnt == 0 || n < l->first->number) {
insert_start (l, n);
return;
}
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
ptr->next = NULL;
lnode *iter = l->first; /* pointer to iterate list */
while (iter->next && n > iter->next->number) {
iter = iter->next;
}
if (iter->next)
ptr->next = iter->next;
iter->next = ptr;
l->cnt++;
}
只要我们扩展你的列表，就应该保持main函数的清除功能为print列表为你和free所有内存分配给列表当你是完成。可以实现此目的的几个辅助函数可以是：
void prn_list (list l)
{
lnode *ptr = l.first;
int i = 0;
while (ptr)
{
printf(" node[%2d] : %d\n", i++, ptr->number);
ptr = ptr->next;
}
}
void free_list (list l)
{
lnode *ptr = l.first;
while (ptr)
{
lnode *del = ptr;
ptr = ptr->next;
free (del);
del = NULL;
}
}
以类似方式删除作品。总而言之，您将获得一个具有输入功能的半稳健列表。请注意，struct也创建了typedefs，减少了输入并提高了可读性。
#include <stdio.h>
#include <stdlib.h>
// #include <conio.h>
typedef struct lnode
{
int number;
struct lnode *next;
} lnode;
typedef struct
{
int cnt;
lnode *first;
} list;
void insert_end (list *l, int n);
void insert_start (list *l, int n);
void insert_ordered (list *l, int n);
void insert_pos (list *l, int n, int pos);
void prn_list (list l);
void free_list (list l);
int main (void)
{
list lk = { 0, NULL };
int num = 0;
int i = 0;
printf ("\n number of nodes to enter: ");
scanf ("%d", &num);
for (i = 0; i < num; i++)
{
int n = 0;
printf (" enter node[%d]->number: ", i);
scanf("%d", &n);
insert_end (&lk, n);
}
printf ("\n The list contains '%d' nodes.\n", lk.cnt);
printf ("\n The list nodes are:\n\n");
prn_list (lk);
printf ("\n enter number to add at start: ");
scanf("%d", &num);
insert_start (&lk, num);
printf ("\n The list contains '%d' nodes.\n", lk.cnt);
printf ("\n The list nodes are:\n\n");
prn_list (lk);
printf ("\n enter number to add in order: ");
scanf("%d", &num);
insert_ordered (&lk, num);
printf ("\n The list contains '%d' nodes.\n", lk.cnt);
printf ("\n The list nodes are:\n\n");
prn_list (lk);
printf ("\n enter number to add at position: ");
scanf("%d", &num);
printf ("\n position must be (0 <= pos <= %d)\n", lk.cnt);
printf ("\n enter position in list for '%d': ", num);
scanf("%d", &i);
insert_pos (&lk, num, i);
printf ("\n The list contains '%d' nodes.\n", lk.cnt);
printf ("\n The list nodes are:\n\n");
prn_list (lk);
printf ("\n Freeing list memory:\n\n");
free_list (lk);
//getch();
return 0;
}
/* insert node at end of list */
void insert_end (list *l, int n)
{
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
ptr->next = NULL;
if (l->cnt == 0)
{
l->first = ptr;
l->cnt++;
return;
}
lnode *iter = l->first; /* pointer to iterate list */
while (iter->next) iter = iter->next;
iter->next = ptr;
l->cnt++;
}
/* insert node at beginning of list */
void insert_start (list *l, int n)
{
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
if (l->cnt == 0)
ptr->next = NULL;
else
ptr->next = l->first;
l->first = ptr;
l->cnt++;
}
/* insert node at end of list */
void insert_ordered (list *l, int n)
{
/* if first node of n < first->number */
if (l->cnt == 0 || n < l->first->number) {
insert_start (l, n);
return;
}
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
ptr->next = NULL;
lnode *iter = l->first; /* pointer to iterate list */
while (iter->next && n > iter->next->number)
iter = iter->next;
if (iter->next)
ptr->next = iter->next;
iter->next = ptr;
l->cnt++;
}
/* insert node at position */
void insert_pos (list *l, int n, int pos)
{
/* validate position */
if (pos < 0 || pos > l->cnt) {
fprintf (stderr, "%s() error: invalid position.\n", __func__);
return;
}
/* if pos 0, insert_start */
if (l->cnt == 0 || pos == 0) {
insert_start (l, n);
return;
}
struct lnode *ptr = NULL;
if (!(ptr = calloc (1, sizeof *ptr))) {
fprintf (stderr, "%s() error: memory exhausted.\n", __func__);
exit (EXIT_FAILURE);
}
ptr->number = n;
ptr->next = NULL;
lnode *iter = l->first; /* pointer to iterate list */
while (--pos)
iter = iter->next;
if (iter->next)
ptr->next = iter->next;
iter->next = ptr;
l->cnt++;
}
/* print all nodes in list */
void prn_list (list l)
{
lnode *ptr = l.first;
int i = 0;
while (ptr)
{
printf(" node[%2d] : %d\n", i++, ptr->number);
ptr = ptr->next;
}
}
/* free memory for all nodes */
void free_list (list l)
{
lnode *ptr = l.first;
while (ptr)
{
lnode *del = ptr;
ptr = ptr->next;
free (del);
del = NULL;
}
}
使用/输出
$ ./bin/ll_single_ins
number of nodes to enter: 3
enter node[0]->number: 5
enter node[1]->number: 7
enter node[2]->number: 9
The list contains '3' nodes.
The list nodes are:
node[ 0] : 5
node[ 1] : 7
node[ 2] : 9
enter number to add at start: 2
The list contains '4' nodes.
The list nodes are:
node[ 0] : 2
node[ 1] : 5
node[ 2] : 7
node[ 3] : 9
enter number to add in order: 6
The list contains '5' nodes.
The list nodes are:
node[ 0] : 2
node[ 1] : 5
node[ 2] : 6
node[ 3] : 7
node[ 4] : 9
enter number to add at position: 4
position must be (0 <= pos <= 5)
enter position in list for '4': 4
The list contains '6' nodes.
The list nodes are:
node[ 0] : 2
node[ 1] : 5
node[ 2] : 6
node[ 3] : 7
node[ 4] : 4
node[ 5] : 9
Freeing list memory:
valgrind内存错误检查
$ valgrind ./bin/ll_single_ins
==22898== Memcheck, a memory error detector
==22898== Copyright (C) 2002-2012, and GNU GPL'd, by Julian Seward et al.
==22898== Using Valgrind-3.8.1 and LibVEX; rerun with -h for copyright info
==22898== Command: ./bin/ll_single_ins
==22898==
number of nodes to enter: 3
enter node[0]->number: 5
enter node[1]->number: 7
enter node[2]->number: 9
The list contains '3' nodes.
<snip>
==22519== HEAP SUMMARY:
==22519== in use at exit: 0 bytes in 0 blocks
==22519== total heap usage: 5 allocs, 5 frees, 80 bytes allocated
==22519==
==22519== All heap blocks were freed -- no leaks are possible
==22519==
==22519== For counts of detected and suppressed errors, rerun with: -v
==22519== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)