Question

我有以下算法用于反转链表。

typedef struct node
{
    int data;
    struct node* next;
}LL_t;

void insert(LL_t** head, int value) 
{
    LL_t* new_node = (LL_t*) malloc(sizeof(LL_t)); 
    new_node->data  = value; 
    new_node->next = (*head); 
    (*head)    = new_node;
}

LL_t* head;
// Post: L contains the data in the original L, but in the reverse order.
void reverse(LL_t * L) {
    LL_t *current, *prev, *next;
    current = head;
    prev = NULL;
    while ( current != NULL )
    {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;  
    }   
    head = prev;
}

int main ()
{
struct node* L = NULL;
insert( &L, 42 );
insert( &L, 36 );
insert( &L, 14 );
insert( &L, 17 );
insert( &L, 48 );
insert( &L, 36 );

print(L);
printf("\n");
reverse(L);
print(L);

return 0;
}

在反向功能之前和之后打印时，列表显示为相同。我相信我不小心使用了局部变量来修改我的列表，而我想直接修改列表（L）。输出：

36 48 17 14 36 42
36 48 17 14 36 42

如何使用全局变量直接修改链表？

Answer 1

您的代码的问题是您将指针值传递给函数reverse并且指针l仍指向第一个元素，而不是指向退出反向函数后的最后一个元素，你需要通过引用传递指针并更改列表的头部，如下所示：

void reverse(LL_t **L) {
    LL_t *current, *prev, *next;
    current = *L;
    prev = NULL;
    while ( current != NULL )
    {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;  
    }
    head = prev;   
    *L = head;
}

应该是

reverse(&L);

进行此更改，现在调用

print(L)并看到您获得了反向链接列表。

如果您有一个固定的原型，就像您在评论中所说的那样，那么请选择

void reverse(LL_t *L) {
    LL_t *current, *prev, *next;
    current = L;
    prev = NULL;
    while ( current != NULL )
    {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;  
    }   
    head = prev;
}

电话应该是

reverse(L);

然后打印电话

print(head);

Answer 2

head = L;//you don't assign to head.
reverse(L);//unused L at reverse.
print(head);

链接列表 - 全局v。局部变量

2 个答案: