为了方便链表的增删操作，一些算法生成一个虚拟头节点，方便统一删除头节点和其他节点。下面代码实现 了删除链表中值为 val 的节点，横线上应填的最佳代码是( )。
struct LinkedNode {
int val;
LinkedNode* next;
LinkedNode(int val):val(val), next(nullptr){}
};
void removeElements(LinkedNode* head, int val) {
if (head == nullptr) {
return;
}
LinkedNode* cur;
LinkedNode* dummyHead = new LinkedNode(0); //虚拟头节点
________________________________ // 在此处填入代码
while(cur ->next ！= nullptr) {
if(cur->next->val == val) {
LinkedNode* tmp = cur->next;
cur->next = cur->next->next;
delete tmp;
tmp = nullptr;
}
else {
cur = cur ->next;
}
}
head = dummyHead->next;
delete dummyHead;
dummyHead = nullptr;
}
- A. dummyHead->next = head; cur = dummyHead;
- B. dummyHead->next = head->next; cur = dummyHead;
- C. dummyHead->next = head; cur = dummyHead->next;
- D. dummyHead->next = head->next; cur = dummyHead->next;
正确答案：A