我正在研究从链表中删除节点的迭代删除功能,我认为代码应该可以正常工作 . 但是当我无法使用“删除”删除列表的第一个节点时 . 代码是:#include using namespace std;
struct Node
{
int data;
Node* next;
};
Node* GetNewNode(int data)
{
Node* newNode = new Node;
newNode->data = data;
newNode->next = NULL;
return newNode;
}
Node* Insert(Node *root, int data)
{
if (root == NULL)
{
root = GetNewNode(data);
}
else
root->next = Insert(root->next, data);
return root;
}
void Delete_k(Node *root, int k)
{
int i = 0;
Node* P = new Node;
if (k == 1)
{
P = root;
root = root->next;
delete P;
}
else
{
for (int i = 1; i <= k - 2; i++)
{
root = root->next;
}
root->next = root->next->next;
}
}
void Output(Node* root)
{
if (root == NULL)
{
root = root->next;
}
while (root != NULL)
{
cout << root->data << " ";
root = root->next;
}
}
int main()
{
int n, a, pos;
Node* root = NULL;
cout << "Input your list hear: ";
cin >> n;
while (n > 0)
{
root = Insert(root, n);
cin >> n;
}
Output(root);
cout << "\nDelete Pos?: ";
cin >> pos;
Delete_k(root, pos);
Output(root);
}
我有这个问题
void Delete_k(Node *root, int k)
{
int i = 0;
Node* P = new Node;
if (k == 1)
{
P = root;
root = root->next;
delete P;
}
else
{
for (int i = 1; i <= k - 2; i++)
{
root = root->next;
}
root->next = root->next->next;
}
}
1 回答
问题:
root的值是通过引用传递的,但指向它的指针不是 .
结果:
Delete_k
的root是main
's root. Delete_K'的root被重新命名和删除的副本 . Main的root现在指向垃圾内存 . 结束游戏计划 .解:
提供对根指针的引用,以便不会复制它 .
或者从Delete_k返回root .