# include <iostream>
using namespace std;
class Node
{
public:
int d;Node*temp1;
Node*next;Node*temp2;
};
void insert(Node*&head,int x)
{
Node*node = new Node(); // allocate memory 2 node let node be an abstract data
node->d = x; // define data in the new node as new data (saving data define in there)
node->next = head; // Let next of the new node as head
head = node; // let pointer name head point new node
}
void print(Node*node)
{
while (node != NULL)
{
cout<<' '<<node->d;
node = node->next;
}
}
void Delete(Node*&head,int n) // Delete node at position
{
int i;Node*node=head;// temp1 points 2(n-1)th
if(n==1)
{
head = node->next; // head now points 2 second node.
return;
}
for(i=0;i<n-2;i++)
{
head = node->next;
} // temp1 points 2 (n-1)th Node
Node*nnode= node->next; // nth node temp1=node temp2=nnode
node-> next = nnode->next; //(n+1)th Node
}
int main()
{
Node*head = NULL; // Start with empty List
int a,n,i,x;
cin>>n;
for(i=0;i<n;i++)
{
cin>>x;
insert(*&head,x);
}
cout<<"Enter a position:";
cin>>a;
Delete(head,a);print(head);
}
输出为:
3 // how many number that singly linked list can received
1 2 3 // define many numbers
Enter a position : 1
2 1 // false output it should be 2 3
输出应为:
3
1 2 3
Enter a position : 1
Linked List is 1->2->3
position 1 is remove // at any position we want 2 remove it will show that position we remove
2->3
Enter a position : 4
No data at 4th position
Linked List is 2->3
在delete函数中，您有一个循环
for(i=0;i<n-2;i++)
{
head = node->next;
}
因为您通过引用传递head，所以您使用此循环主动销毁列表。 此外，由于前面有node=head，因此在第一次迭代中，分配实际上是head=head->next。
您需要使用变量node而不是head:
for(i=0;i<n-2;i++)
{
node = node->next;
}
您还需要防止超出列表的末尾:
for(i = 0; (i < n - 2) && (node->next != nullptr) ;i++)
对于初学者，单个链表节点的声明具有多余的数据成员temp1和temp2，这是没有意义的。
声明可以看起来像
struct Node
{
int data;
Node *next;
};
在本例中，函数insert（您可以像这样调用
insert(head,x);
而不是
insert(*&head,x);
正如您正在做的）将看起来像
void insert( Node * &head, int x )
{
head = new Node { x, head };
}
在C++（和C）中，索引从0开始。 因此函数delete也应接受从0开始的索引。 相应参数的类型应为无符号整数类型，例如size_t。 否则，用户可以传递一个负数作为索引。
该函数会产生内存泄漏，因为它实际上不释放已分配的节点。 当指向头节点的指针等于NULL时，它可以调用未定义的行为。 而且一般情况下，这个函数是没有意义的。
它可以按以下方式定义
bool Delete( Node * &head, size_t n )
{
Node **current = &head;
while ( *current && n-- )
{
current = &( *current )->next;
}
bool success = *current != nullptr;
if ( success )
{
Node *tmp = *current;
*current = ( *current )->next;
delete tmp;
}
return success;
}
这里有一个演示程序。
#include <iostream>
struct Node
{
int data;
Node *next;
};
void insert( Node * &head, int x )
{
head = new Node { x, head };
}
bool Delete( Node * &head, size_t n )
{
Node **current = &head;
while ( *current && n-- )
{
current = &( *current )->next;
}
bool success = *current != nullptr;
if ( success )
{
Node *tmp = *current;
*current = ( *current )->next;
delete tmp;
}
return success;
}
std::ostream & print( Node * &head, std::ostream &os = std::cout )
{
for ( Node *current = head; current != nullptr; current = current->next )
{
os << current->data << " -> ";
}
return os << "null";
}
int main()
{
Node *head = nullptr;
for ( int i = 3; i != 0; i-- ) insert( head, i );
print( head ) << '\n';
size_t pos = 0;
if ( Delete( head, pos ) )
{
print( head ) << '\n';
}
else
{
std::cout << "No data at the position " << pos << '\n';
}
pos = 4;
if ( Delete( head, pos ) )
{
print( head ) << '\n';
}
else
{
std::cout << "No data at the position " << pos << '\n';
}
pos = 1;
if ( Delete( head, pos ) )
{
print( head ) << '\n';
}
else
{
std::cout << "No data at the position " << pos << '\n';
}
pos = 0;
if ( Delete( head, pos ) )
{
print( head ) << '\n';
}
else
{
std::cout << "No data at the position " << pos << '\n';
}
return 0;
}
其输出为
1 -> 2 -> 3 -> null
2 -> 3 -> null
No data at the position 4
2 -> null
null
