A20877. $下面函数用“哑结点”统一处理删除单向链表中的头结点与中间结点。横线处应填（）。struct Node{ int val; Node* next; Node(int v):val(v),next(nullptr){} }; Node* eraseAll(Node* head, int x){ Node dummy(0); dummy.next = head; Node* cur = &dummy\dots$

单选题困难

知识点

链表删除哑结点指针操作

题目描述

下面函数用“哑结点”统一处理删除单向链表中的头结点与中间结点。横线处应填（ ）。 struct Node{ int val; Node* next; Node(int v):val(v),next(nullptr){} }; Node* eraseAll(Node* head, int x){ Node dummy(0); dummy.next = head; Node* cur = &dummy; while(cur->next){ if(cur->next->val == x){ Node* del = cur->next; ----------------------- delete del; }else cur = cur->next; } return dummy.next; }

cur = cur->next;

cur->next = del->next;

del->next = cur->next;

cur->next = nullptr;

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