我试图交换链表中节点的位置 然后使用sort函数进行排序。我在其中一个函数中遇到逻辑错误。当我运行程序时，它会进入无限循环。
更新代码
int adjuctNodes(struct student_record_node** n1, struct student_record_node** n2)
{
struct student_record_node* prev_;
struct student_record_node* next_;
return((*n1)->next_==(*n2) && (*n2)->prev_ == (*n1))||
( (*n1)->prev_ ==(*n2) && (*n2)->next_ == (*n1) );
}
void updateOuterPointer(struct student_record_node** n)
{
struct student_record_node* next_;
struct student_record_node* prev_;
if((*n)->next_!=NULL)
(*n)->prev_->next_=(*n);
if((*n)->next_ !=NULL)
(*n)->next_->prev_=(*n);
}
/*Swaping */
void swap(struct student_record_node** node1, struct student_record_node** node2)
{
struct student_recod_node* prev_;
struct student_recod_node* next_;
struct student_record_node* ptr=(*node1)->next_;
if(adjucntNodes((node1),(node2)))
{
(node1)->prev_=pnode2;
(node2)->prev_=pnode0;
(node1)->next_=pnode3;
(node2)->next_=pnode1;
}else{
(node1)->prev_=pnode1;
(node2)->prev_=pnode0;
(node1)->next_=pnode3;
(node2)->next_=pnode2;
}
updateOuterPointer((node1));
updateOuterPointer((node2));
}
/*Sorting linked list*/
void sort(struct student_record_node**recordsHead,int(*compare_fcn)(struct
student_record_node*,struct student_record_node*))
{
int swapped;
struct student_record_node *ptr1=*recordsHead;
struct student_record_node *lptr = NULL;
if (ptr1 == NULL)
return;
do
{
swapped = 0;
ptr1 = *recordsHead;
while (ptr1->next_ != lptr)
{
if (compare_fcn(ptr1,ptr1->next_))
{
printf("swapping\n");
swap(&ptr1,&ptr1->next_);
if(ptr1==*recordsHead)
{
(*recordsHead)=ptr1->next_;
}
swapped=1;
}
else ptr1 = ptr1->next_;
}
lptr = ptr1;
;
}
while (swapped);
}
答案 0 :(得分：1)
要处理节点与公共代码相邻或不相邻的情况，首先将（外部）指针交换到两个节点，然后交换两个节点的（内部）指针。如果节点是相邻的，这将最终根据需要旋转指针，并且如果节点不相邻则交换指针对。注意节点是否相邻，其中一个是＆＃34;外部＆＃34;指针将成为＆＃34;其他＆＃34;节点内部指针，反之亦然，但它仍然有效：swap＆＃34; external＆＃34;指针首先，然后＆＃34;内部＆＃34;指针下一步。
确保在执行交换时根据需要使用临时指针（技术上指向节点指针的指针），否则在交换操作的一部分中覆盖节点指针。如果卡住，我可以稍后用一个例子更新。
update - 图表类型示例以显示发生的情况，使用单个链接列表仅使用下一个指针作为示例。假设您从5个节点开始，0到4：
0->1->2->3->4
0->3
2->1
然后交换1-＆gt;和3-＆gt;
1->4
3->2
导致
0->3->2->1->4
从0-> 1-> 2-> 3-> 4交换1和2,0-> 0开始。并且1->是外部的，1->和2->是内部的。交换0->和1-＆gt;
0->2
1->1
然后交换1-＆gt;和2->
1->3
2->1
导致
0->2->1->3->4
示例交换代码。此代码假定有指向第一个节点的头指针，以及指向最后一个节点（或NULL）的尾指针。
struct student_record_node *Head = &firstnode; /* head */
struct student_record_node *Tail = &lastnode; /* tail (NULL is ok) */
/* swap function */
void swap(struct student_record_node **Head,
struct student_record_node **Tail,
struct student_record_node *node1,
struct student_record_node *node2)
{
struct student_record_node **en1 /* & external next ptr to 1 */
struct student_record_node **en2 /* & external next ptr to 2 */
struct student_record_node **ep1 /* & external prev ptr to 1 */
struct student_record_node **ep2 /* & external prev ptr to 2 */
struct student_record_node *tmp /* temp node ptr */
en1 = (node1->prev_ != NULL) ? &(node1->prev_->next_) : Head;
en2 = (node2->prev_ != NULL) ? &(node2->prev_->next_) : Head;
ep1 = (node1->next_ != NULL) ? &(node1->next_->prev_) : Tail;
ep2 = (node2->next_ != NULL) ? &(node2->next_->prev_) : Tail;
/* swap en1, en2 */
tmp = *en1;
*en1 = *en2;
*en2 = tmp;
/* swap ep1, ep2 */
tmp = *ep1;
*ep1 = *ep2;
*ep2 = tmp;
/* swap n1, n2 next_ */
tmp = node1->next_;
node1->next_ = node2->next_;
node2->next_ = tmp;
/* swap n1, n2 prev_ */
tmp = node1->prev_;
node1->prev_ = node2->prev_;
node2->prev_ = tmp;
}
/* call to swap function */
swap(&Head, &Tail, node1, node2);
答案 1 :(得分：1)
原始代码中存在两个主要问题，可能还有第三个问题：
swap函数不起作用，但sort函数只交换相邻节点！ptr1和ptr1->next_后，sort函数检查交换的第一个节点ptr1是否位于列表的头部，如果所以使ptr1->next_成为列表的头部。但是，这两个节点现在的顺序相反，所以在这种情况下它应该使ptr1->prev_成为列表的头部。sort函数当前似乎期望比较函数返回0。这可能是也可能不是错误，但它是非常规的。此外，可以简化swap函数的接口，因为不需要将指针传递给指向节点的指针。
以下示例程序修复了上述问题：
#include <stdio.h>
#include <string.h>
struct student_record_node {
struct student_record_node *next_;
struct student_record_node *prev_;
const char *name;
unsigned int age;
};
void swap(struct student_record_node *node1, struct student_record_node *node2)
{
struct student_record_node *ptr1, *ptr2;
/* Swap the 'next_' pointers, taking adjacency into account. */
ptr1 = node1 == node2->next_ ? node2 : node2->next_;
ptr2 = node2 == node1->next_ ? node1 : node1->next_;
node1->next_ = ptr1;
node2->next_ = ptr2;
/* Swap the 'prev_' pointers, taking adjacency into account. */
ptr1 = node1 == node2->prev_ ? node2 : node2->prev_;
ptr2 = node2 == node1->prev_ ? node1 : node1->prev_;
node1->prev_ = ptr1;
node2->prev_ = ptr2;
/* Fix the links from other nodes. */
if (node1->next_) node1->next_->prev_ = node1;
if (node1->prev_) node1->prev_->next_ = node1;
if (node2->next_) node2->next_->prev_ = node2;
if (node2->prev_) node2->prev_->next_ = node2;
}
int compare_ages(const struct student_record_node *a,
const struct student_record_node *b)
{
return a->age < b->age ? -1 : a->age > b->age ? 1 : 0;
}
int compare_names(const struct student_record_node *a,
const struct student_record_node *b)
{
return strcmp(a->name, b->name);
}
void sort(struct student_record_node **recordsHead,
int(*compare_fcn)(const struct student_record_node *,
const struct student_record_node *))
{
int swapped;
struct student_record_node *ptr1;
struct student_record_node *lptr = NULL;
if (*recordsHead == NULL)
return;
do
{
swapped = 0;
ptr1 = *recordsHead;
while (ptr1->next_ != lptr)
{
if (compare_fcn(ptr1, ptr1->next_) > 0)
{
printf("swapping (%s:%u <=> %s:%u)\n", ptr1->name, ptr1->age,
ptr1->next_->name, ptr1->next_->age);
swap(ptr1, ptr1->next_);
if (ptr1 == *recordsHead)
{
/* ptr1 is now the second node. */
(*recordsHead) = ptr1->prev_;
}
swapped = 1;
}
else
{
ptr1 = ptr1->next_;
}
}
lptr = ptr1;
}
while (swapped);
}
void dump(const struct student_record_node *students)
{
const struct student_record_node *prev_ = NULL;
unsigned int pos = 0;
while (students)
{
if (students->prev_ != prev_)
{
printf("[%u] ** Bad prev_ link!\n", pos);
}
printf("[%u] %s:%u\n", pos, students->name, students->age);
pos++;
prev_ = students;
students = students->next_;
}
}
int main(void)
{
static struct student_record_node testdata[] =
{
{ testdata + 1, NULL, "susan", 20 },
{ testdata + 2, testdata + 0, "bill", 21 },
{ testdata + 3, testdata + 1, "joe", 18 },
{ testdata + 4, testdata + 2, "tom", 19 },
{ NULL, testdata + 3, "karen", 21 },
};
struct student_record_node *students = testdata;
puts("Original order:");
dump(students);
puts("By name:");
sort(&students, compare_names);
dump(students);
puts("By age:");
sort(&students, compare_ages);
dump(students);
return 0;
}
输出：
Original order:
[0] susan:20
[1] bill:21
[2] joe:18
[3] tom:19
[4] karen:21
By name:
swapping (susan:20 <=> bill:21)
swapping (susan:20 <=> joe:18)
swapping (tom:19 <=> karen:21)
swapping (susan:20 <=> karen:21)
[0] bill:21
[1] joe:18
[2] karen:21
[3] susan:20
[4] tom:19
By age:
swapping (bill:21 <=> joe:18)
swapping (karen:21 <=> susan:20)
swapping (karen:21 <=> tom:19)
swapping (bill:21 <=> susan:20)
swapping (bill:21 <=> tom:19)
swapping (susan:20 <=> tom:19)
[0] joe:18
[1] tom:19
[2] susan:20
[3] bill:21
[4] karen:21