大佬们，不是说链表当作函数参数时，想要改变原链表的值需要使用引用传递吗，
为什么我这个delete_last_node函数能删掉原链表的最后一个节点呀。。。
但是上边的insertHead必须使用引用传递才能给原链表添加新的头节点。。

typedef struct ListNode{
int val;
struct ListNode *next;
ListNode(int x) : val(x), next(NULL){};
};

void insertHead(ListNode *&head, int x){
ListNode *newNode = new ListNode(x);
newNode->next = head;
head = newNode;
}

void delete_last_node(ListNode *head){
ListNode *p1 = head;
ListNode *p2 = head->next;
while (p2->next != NULL) {
p1 = p2;
p2 = p2->next;
}
p1->next = NULL;
delete p2;
}

void clear(ListNode *head){
if (head == NULL)
return;
ListNode *cur = head->next;
while (cur != NULL) {
head->next = cur->next;
delete cur;
cur = head->next;
}
delete head;
}

void print_listnode(ListNode *head){
ListNode *cur = head;
while (cur != NULL) {
std::cout << cur->val << "->";
cur = cur->next;
}
std::cout << std::endl;
}

int main(){
ListNode *node = new ListNode(0);
node->next = new ListNode(1);
node->next->next = new ListNode(2);
insertHead(node, 3);
delete_last_node(node);
print_listnode(node);
clear(node);
system("pause");
return 0;
}

等等，我好像明白了

指针一样是操作的本体。

*&是什么语法，你写的c还是c++啊

抄老师的代码

没老师就抄百度，百度上我也看过，写的比你好一点