我正在处理一个链表,我想将列表中的一个节点从一个位置移动到另一个位置,而不会弄乱任何内容。
链表的结构是:
struct Frame
{
char* name; // Name and path are just pointers to strings
unsigned int duration;
char* path;
};
typedef struct Frame frame_t;
和
struct Link
{
frame_t *frame;
struct Link *next;
};
(这就是我被要求做的,而不是我的选择)
现在我需要的是一个函数,它接收链表和一个字符串(一个节点的名称)和一个整数,然后将具有该名称的节点移动到该位置(接收的整数)第一个位置为1,而不是0(这就是所要求的)
所以例如: 如果列表包含节点:[pic1,pic2,pic3,pi4] 并且用户请求将“pic1”移动到pos 3,然后新列表将是: [pic2,pic3,pic1,pic4](而pic2将是新的头部)
我尝试了一些版本,但他们总是只有80%工作(切断列表或不移动到正确的位置)。有什么想法吗?
继承了我尝试过的功能:
void changePos(link_t** anchor_link, char* name1, int pos)
{
link_t* currLink = *anchor_link;
link_t* temp = NULL;
link_t* temp2 = NULL;
int i;
if (strcmp(name1, currLink->frame->name) == 0 && currLink->next)
{
*anchor_link = (*anchor_link)->next;
temp = currLink;
}
else
{
while (strcmp(name1, currLink->next->frame->name) != 0 && currLink->next)
{
currLink = currLink->next;
}
temp = currLink->next;
}
currLink = *anchor_link;
for (i = 1; i < pos - 1; i++) // Go up until the node before the pos (meaning if pos is 4 then node 3)
{
currLink = currLink->next;
}
// Now we insert the temp node at the pos
temp2 = currLink->next->next;
currLink->next->next = temp;
temp->next = temp2;
}
答案 0 :(得分:0)
我认为您的一般方法很好:找到目标节点,将其从列表中删除,然后将其重新插入指定位置。但是,您的代码存在一些问题。
首先,考虑这个while()
循环的条件:
while (strcmp(name1, currLink->next->frame->name) != 0 && currLink->next)
如果您不确定currLink->next
是NULL
(并且您应该是),那么您必须在解除引用之前执行空检查,您在表达式中执行此操作{ {1}}。实际上,如果代码到达列表的末尾而未找到指定的名称,则代码会显示未定义的行为。
其次,你的变量命名有点不尽如人意。什么&#39; currLink->next->frame->name
?什么&#39; temp
?鉴于temp2
没有1
,name1
中的name2
表示什么?以明确传达其目的的方式命名变量有助于编写和分析代码。例如,&#34; target_node&#34;有什么问题?
第三,您的插入代码错误:
temp2 = currLink->next->next; currLink->next->next = temp; temp->next = temp2;
您打算在*temp
和*currLink
之间插入*currLink->next
,但实际上您已将放在 *currLink->next
之后,然后关闭两个 - 元素循环。此外,我怀疑使用temp2
只是混淆了这段代码。我会写这个:
temp->next = currLink->next->next;
currLink->next = temp;
第四,您没有在列表中插入第1位的条款。你做这件事的方式需要一个特殊的例子。
答案 1 :(得分:0)
这是我提出的代码 - changePos()
函数和我创建的测试代码,以确保它是正确的。
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Frame
{
char *name;
unsigned int duration; // Effectively unused
// char *path; // Actually unused
};
typedef struct Frame Frame;
struct Link
{
Frame *frame;
struct Link *next;
};
typedef struct Link Link;
static void print_list(Link *root);
static void changePos(Link **anchor_link, const char *name, int pos)
{
assert(anchor_link != 0 && name != 0 && pos >= 0);
Link *root = *anchor_link;
Link *link = root;
Link *prev = 0;
int count = 0;
while (link != 0 && strcmp(link->frame->name, name) != 0)
{
prev = link;
link = link->next;
count++;
}
if (link == 0) // Name not found - no swap!
return;
if (count == pos) // Already in target position - no swap
return;
if (count == 0) // Moving first item; update root
{
assert(link == root);
*anchor_link = root->next;
root = *anchor_link;
}
else
{
assert(prev != 0);
prev->next = link->next;
}
// link is detached; now where does it go?
if (pos == 0) // Move to start; update root
{
link->next = root;
*anchor_link = link;
return;
}
Link *node = root;
for (int i = 0; i < pos - 1 && node->next != 0; i++)
node = node->next;
link->next = node->next;
node->next = link;
}
static void print_list(Link *root)
{
const char *pad = "";
while (root != 0)
{
printf("%s[%s]", pad, root->frame->name);
root = root->next;
pad = "->";
}
}
static void free_frame(Frame *frame)
{
if (frame != 0)
{
free(frame->name);
free(frame);
}
}
static void free_link(Link *link)
{
while (link != 0)
{
Link *next = link->next;
free_frame(link->frame);
free(link);
link = next;
}
}
static Frame *make_frame(const char *name, unsigned int number)
{
Frame *frame = malloc(sizeof(*frame));
if (frame != 0)
{
frame->name = strdup(name);
frame->duration = number;
}
return frame;
}
static Link *make_link(const char *name, unsigned int number)
{
Link *link = malloc(sizeof(*link));
if (link != 0)
{
link->frame = make_frame(name, number);
link->next = 0;
}
return link;
}
static Link *make_list(int num, Frame *frames)
{
Link *head = 0;
Link *tail = 0;
for (int k = 0; k < num; k++)
{
Link *link = make_link(frames[k].name, frames[k].duration);
assert(link != 0 && link->frame != 0); // Lazy!
if (head == 0)
head = link;
if (tail != 0)
tail->next = link;
tail = link;
}
return head;
}
int main(void)
{
Frame frames[] =
{
{ "pic0", 0 },
{ "pic1", 1 },
{ "pic2", 2 },
{ "pic3", 3 },
{ "pic4", 4 }, // Never in the list, but searched for
};
enum { NUM_FRAMES = sizeof(frames) / sizeof(frames[0]) };
for (int i = 0; i < NUM_FRAMES; i++)
{
for (int j = 0; j < NUM_FRAMES; j++)
{
Link *head = make_list(NUM_FRAMES - 1, frames);
print_list(head);
printf(" == %s to %u == ", frames[i].name, j);
changePos(&head, frames[i].name, j);
print_list(head);
putchar('\n');
free_link(head);
}
}
return 0;
}
我怀疑changePos()
中的代码可以简化一点,但我还没有发现。这是一个与数字位置和列表的丑陋接口 - 使用另一个名称来识别节点应该移动到的位置会更自然。对于数字位置,您很想使用数组而不是列表。
示例输出:
[pic0]->[pic1]->[pic2]->[pic3] == pic0 to 0 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic0 to 1 == [pic1]->[pic0]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic0 to 2 == [pic1]->[pic2]->[pic0]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic0 to 3 == [pic1]->[pic2]->[pic3]->[pic0]
[pic0]->[pic1]->[pic2]->[pic3] == pic0 to 4 == [pic1]->[pic2]->[pic3]->[pic0]
[pic0]->[pic1]->[pic2]->[pic3] == pic1 to 0 == [pic1]->[pic0]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic1 to 1 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic1 to 2 == [pic0]->[pic2]->[pic1]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic1 to 3 == [pic0]->[pic2]->[pic3]->[pic1]
[pic0]->[pic1]->[pic2]->[pic3] == pic1 to 4 == [pic0]->[pic2]->[pic3]->[pic1]
[pic0]->[pic1]->[pic2]->[pic3] == pic2 to 0 == [pic2]->[pic0]->[pic1]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic2 to 1 == [pic0]->[pic2]->[pic1]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic2 to 2 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic2 to 3 == [pic0]->[pic1]->[pic3]->[pic2]
[pic0]->[pic1]->[pic2]->[pic3] == pic2 to 4 == [pic0]->[pic1]->[pic3]->[pic2]
[pic0]->[pic1]->[pic2]->[pic3] == pic3 to 0 == [pic3]->[pic0]->[pic1]->[pic2]
[pic0]->[pic1]->[pic2]->[pic3] == pic3 to 1 == [pic0]->[pic3]->[pic1]->[pic2]
[pic0]->[pic1]->[pic2]->[pic3] == pic3 to 2 == [pic0]->[pic1]->[pic3]->[pic2]
[pic0]->[pic1]->[pic2]->[pic3] == pic3 to 3 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic3 to 4 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic4 to 0 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic4 to 1 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic4 to 2 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic4 to 3 == [pic0]->[pic1]->[pic2]->[pic3]
[pic0]->[pic1]->[pic2]->[pic3] == pic4 to 4 == [pic0]->[pic1]->[pic2]->[pic3]
输出的LHS是之前的列表 - 总是相同的,在序列pic0到pic3中。输出的RHS是调用changePos
后的列表。对于位置 n = 0..3 ,您可以看到'picn'依次从位置0迁移到位置3。名称pic4
永远不会插入到列表中,因此在搜索时不会发生任何变化。此外,当您尝试将任何名称移动到不存在的位置时,它将移动到列表中的最后位置。小于0的位置被声明无效。
代码有偶然的错误检查。它断言内存分配问题以及其他一些问题(例如小于0的位置)。
在changePos()
中写下任何有价值的内容之前,我使用虚拟无操作版本changePos()
干净地运行了线束。
使用我的习惯性编译器警告选项,代码在Mac OS X 10.11.5上使用GCC 6.1.0和Valgrind 3.12.0.SVN进行编译和运行:
$ gcc -O3 -g -std=c11 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes \
> -Wold-style-definition -Werror so.3807-9550.c -o so.3807-9550
$