链表将节点移动到C端

Question

我正试图将所选元素（节点）从链接列表移动到此列表的末尾。

struct* move(struct* list , struct* element)
{
  while(list->next != NULL)
   {
      list = list->next;
   }
  list = element;
  return list;
}

首先，结构体element是list的一部分，但我需要将它移到末尾。然而，他的doenst似乎在工作。。。

Answer 1

当您执行list = element;时，实际上会使list指向element，这不会更改您的列表。

你想做的是：

list->next = element;
element->next = NULL;

但是您仍然需要将element从其先前的位置中删除。move()的样本如下：

struct struct_type * move(struct struct_type * list , struct struct_type * element) {
    while (list->next != NULL) {
        if (list->next == element)
            list->next = list->next->next;
        list = list->next;
    }
    list->next = element;
    element->next = NULL;
    return list;    

}

如注释中所述，您必须指定列表的结构类型。还要注意，返回list会返回列表中除一个元素之外的最后一个元素（我认为这不是预期的行为）。

EDIT：处理第一个元素并返回列表的顶部元素。

struct struct_type * move(struct struct_type * list , struct struct_type * element) {
    struct struct_type *l = list; //keeps reference to the list
    if (l == NULL)
        return l;
    if (l == element)
        list = l->next;
    while (l->next != NULL) {
        if (l->next == element)
            l->next = l->next->next;
        l = l->next;
    }
    l->next = element;
    element->next = NULL;
    return list;    
}

Answer 2

#include <stdio.h>
struct node {
    char name;
    struct node *next;
};
struct node *move(struct node *list, struct node *element){
    struct node *head = list;
    struct node *prev = NULL, *curr = head;
    if(list == NULL || list->next == NULL)
        return head;
    while(curr){
        if(curr == element){
            if(curr == head){
                head = curr->next;
            } else {
                prev ->next = curr->next;
            }
            curr = curr->next;
        } else {
            prev = curr;
            curr = curr->next;
        }
    }
    prev->next = element;
    element->next = NULL;
    return head;
}
void print(struct node *list);
int main(void) {
    struct node a[] = {
        {'A', NULL},
        {'B', NULL},
        {'C', NULL},
        {'D', NULL},
        {'E', NULL}
    }, *head;
    a[0].next = &a[1];
    a[1].next = &a[2];
    a[2].next = &a[3];
    a[3].next = &a[4];
    print(head=a);//A->B->C->D->E
    head = move(head, head);
    print(head);//B->C->D->E->A
    head = move(head, &a[2]);
    print(head);//B->D->E->A->C
    head = move(head, &a[2]);
    print(head);//B->D->E->A->C
    return 0;
}
void print(struct node *list){
    while(list){
        printf("%c", list->name);
        list = list->next;
        if(list)
            printf("->");
    }
    printf("n");
}