链表：冒泡排序指针（C）
我正在做一项大学作业。我正在尝试用 C 编写链表排序。我不允许交换值 - 只能交换指针。
这是我的排序功能：
struct node *sort_list(struct node *head) {
bool swapped ;
struct node *cur = head, *first ;
if ( head == NULL || head->next == NULL ) return head ;
else {
do {
swapped = false ;
while ( cur != NULL && cur->next != NULL ){
if (cur->value > cur->next->value){
cur = swap_with_next( cur ) ;
swapped = true ;
}
cur = cur->next ;
}
} while (swapped == true) ;
}
return head ;
}
和交换功能：
struct node *swap_with_next(struct node *n) {
struct node *tmp ;
tmp = n ;
n = n->next ;
tmp->next = n->next ;
n->next = tmp ;
return n ;
}
问题： 输出不正确：
input: 5->2->NULL
output: 5->NULL
input: 9->1->5->2->8->3
output: 9->5->8
任何帮助将不胜感激！
I'm working on a university assignment. I am trying to write a linked-list sort in C. I am not allowed to swap values - only pointers.
Here is my sorting function:
struct node *sort_list(struct node *head) {
bool swapped ;
struct node *cur = head, *first ;
if ( head == NULL || head->next == NULL ) return head ;
else {
do {
swapped = false ;
while ( cur != NULL && cur->next != NULL ){
if (cur->value > cur->next->value){
cur = swap_with_next( cur ) ;
swapped = true ;
}
cur = cur->next ;
}
} while (swapped == true) ;
}
return head ;
}
And the swap function:
struct node *swap_with_next(struct node *n) {
struct node *tmp ;
tmp = n ;
n = n->next ;
tmp->next = n->next ;
n->next = tmp ;
return n ;
}
Problem: Incorrect output:
input: 5->2->NULL
output: 5->NULL
input: 9->1->5->2->8->3
output: 9->5->8
Any help would be greatly appreciated!
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评论（2）
好的，假设有 3 个节点 A、B 和 C（按顺序）。您正在尝试交换 B 和 C。因此，您将指向 B 的指针传递给
swap函数。现在，所有指针操作都是正确的，但是，您错过了一件事。完成交换后，A 的下一个指针应指向 C。你从来没有设置过这个。当你完成交换函数后，A的next指针仍然指向B，这显然是不正确的。这就是问题所在。Ok, so, let's say there are 3 nodes A, B and C (in that order). You are trying to swap B and C. So, you are passing the pointer to B to the
swapfunction. Now, all the pointer manipulations are correct but, you miss one thing. A's next pointer should point to C when you finish the swap. You never set this. When you finish the swap function, A's next pointer still points to B which is obviously not correct. That's the problem.您正在传回一个很好的指针，但是在交换中您正在操纵指针的副本，而不是您试图更改的“实际”指针。您正在尝试操作参数以在交换内部进行交换，但这只是更改在调用该函数时创建的该节点的副本。如果您想在函数内更改它，则需要参数是指向指针的指针。
You are passing back a pointer which is good, but in swap you are manipulating a copy of the pointer not the "actual" one you are trying to change. You are trying to manipulate the parameter to swap inside of swap but that is only changing a copy of that node that was created on calling the function. You need the parameter to be a pointer to the pointer, if you want to change it inside the function.