我正在努力学习C ++，因为我将不得不学习一门课程，而且我来自Java。我目前正在阅读“跳入C ++”一书并完成练习。阅读关于链表的部分后，它告诉我创建自己的链表并有一个方法可以删除一个元素（使用练习中的指针）。
到目前为止，我已经能够将值添加到链接列表中，并显示我链接的链接列表。在执行我的remove元素方法之后，让我的程序明确告诉我它删除了特定内存地址的值，我再次显示列表，发现我的值仍然以某种方式出现在应该被删除的内存地址。
这是我的removeElement方法：
// remove an element from the linked list
void removeElement(int remValue) {
// to remove an element, we go through the list, find the value given
// if we find it, stop
// to remove, disconnect the link
// relink the two values now (ie. value 1->2->3->NULL, 2 is removed, 1->3->NULL )
LinkedList* current = head;
LinkedList* next = current;
while(current != NULL) {
if(current->value == remValue) { // if match
break; // break out of while
}
else {
cout << "Value " << current->value << " does not match " << remValue << ".\n";
next = current; // save in case
current = current->pNextValue; // go to next value
}
} // end while
if(current == NULL) { // if we reached end of list
cout << "Can't remove value: no match found.\n"; // no match, cant remove
} else { // found match
cout << "Deleting: " << current << "\n";
delete current;
current = next->pNextValue; // current is updated
}
}
以下是链接列表的完整代码（包括一些测试以查看内容的来源）：
我意识到我的大部分代码效率都不高，而且对于一个真实的链表不常见，我只想弄清楚指针以及如何在最基本的链表中使用它们。
答案 0 :(得分：8)
您实际上取消链接您删除的节点。
您需要跟踪之前的节点，并使其next指针指向当前节点next节点。还要考虑要删除的节点是第一个节点时的特殊情况。
答案 1 :(得分：1)
您需要更新列表中的链接以删除您删除的节点。
请注意，使用delete运算符不会更改内存中的任何值。它只是告诉操作系统您不再在内存中使用该位置，它可以将其回收用于其他用途。
答案 2 :(得分：1)
@Joachim Pileborg是对的，您需要记录上一个节点，而不是next。
试试这段代码：
// remove an element from the linked list
void removeElement(int remValue) {
LinkedList* prev = head; // empty header
LinkedList* current = head->pNextValue; // the first valid node
while(current != NULL) {
if(current->value == remValue) {
break;
}
else {
cout << "Value " << current->value << " does not match " << remValue << ".\n";
prev = current;
current = current->pNextValue; // go to next value
}
}
if(current == NULL) { // if we reached end of list or the list is empty
cout << "Can't remove value: no match found.\n";
} else {
cout << "Deleting: " << current << "\n";
prev->pNextValue = current->pNextValue; // unlink the node you remove
delete current; // delete the node
}
}
答案 3 :(得分：0)
#include <iostream>
#include <cstdlib>
using namespace std;
class Node
{
public:
Node* next;
int data;
Node();
~Node();
void print();
};
class LinkedList
{
public:
int length;
Node* head;
LinkedList();
~LinkedList();
void add(int data);
void remove(int data);
Node* search(int data);
void print();
void size();
};
Node::Node(){
//set default values;
}
Node::~Node(){
cout << "NODE DELETED" <<endl;
}
void Node::print(){
Node* node = this;
if(node != NULL){
cout << "===============================" << endl;
cout << this->data << endl;
cout << "===============================" << endl;
}
}
LinkedList::LinkedList(){
this->length = 0;
this->head = NULL;
}
LinkedList::~LinkedList(){
cout << "LIST DELETED" <<endl;
}
void LinkedList::add(int data){
Node* node = new Node();
node->data = data;
node->next = this->head;
this->head = node;
this->length++;
}
void LinkedList::remove(int data){
if(this->length == 0){
cout << "The list is empty" << endl;
}else if(this->head->data == data){
Node* current = head;
this->head = this->head->next;
delete current;
this->length--;
}else{
Node* previous = this->head;
Node* current = head->next;
while(current != NULL) {
if(current->data == data) {
break;
}
else {
previous = current;
current = current->next;
}
}
if(current == NULL) {
cout << "Can't remove value: no match found.\n";
} else {
previous->next = current->next;
delete current;
this->length--;
}
}
}
Node* LinkedList::search(int data) {
Node* head = this->head;
while(head){
if(head->data == data){
return head;
}else{
head = head->next;
}
}
cout << "No match found.\n";
return NULL;
}
void LinkedList::print(){
if(this->length == 0){
cout << "The list is empty" << endl;
}else{
Node* head = this->head;
int i = 1;
cout << "===============================" << endl;
while(head){
cout << i << ": " << head->data << endl;
head = head->next;
i++;
}
cout << "===============================" << endl;
}
}
void LinkedList::size(){
cout << "List Length: " << this->length << endl;
}
int main(int argc, char const *argv[])
{
LinkedList* list = new LinkedList();
for (int i = 0; i < 5; ++i)
{
if(i == 3){
list->add(105);
list->add(106);
}
list->add(rand() % 100);
}
list->print();
list->size();
list->remove(105);
list->print();
list->size();
Node* node = list->search(106);
node->print();
delete list;
return 0;
}
答案 4 :(得分：0)
// C++ program to delete a node in
// singly linked list recursively
#include <bits/stdc++.h>
using namespace std;
struct node {
int info;
node* link = NULL;
node() {}
node(int a)
: info(a)
{
}
};
/*
Deletes the node containing 'info' part as val and
alter the head of the linked list (recursive method)
*/
void deleteNode(node*& head, int val)
{
// Check if list is empty or we
// reach at the end of the
// list.
if (head == NULL) {
cout << "Element not present in the list\n";
return;
}
// If current node is the node to be deleted
if (head->info == val) {
node* t = head;
head = head->link; // If it's start of the node head
// node points to second node
delete (t); // Else changes previous node's link to
// current node's link
return;
}
deleteNode(head->link, val);
}
// Utility function to add a
// node in the linked list
// Here we are passing head by
// reference thus no need to
// return it to the main function
void push(node*& head, int data)
{
node* newNode = new node(data);
newNode->link = head;
head = newNode;
}
// Utility function to print
// the linked list (recursive
// method)
void print(node* head)
{
// cout<<endl gets implicitly
// typecasted to bool value
// 'true'
if (head == NULL and cout << endl)
return;
cout << head->info << ' ';
print(head->link);
}
int main()
{
// Starting with an empty linked list
node* head = NULL;
答案 5 :(得分：0)
代码基于 @Some programmer dude 的回答。
void removeElement(int remValue) {
LinkedList * current, next, previous;
current = *&head;
while(current != NULL){
// `next` node
next = current->next;
// if match
if(current->value == remValue){
// if `previous` and `next` has a node, link the `previous` node to `next` node
if(previous != NULL && next != NULL) previous->next = next;
// if `current` is the last node, link `previous` node to `NULL`
else if(previous != NULL && next == NULL) previous->next = NULL;
// if `current` node is the first node, set the header node to `next` node
else if(previous == NULL && next != NULL) *&head = next;
// if `current` node is the only item, empty the list
else *&head = NULL;
cout << "Deleting: " << current->value << "\n";
// delete node
delete current;
// exit the function
return;
// if not match
} else cout << "Value " << current->value << " does not match " << remValue << ".\n";
// Variables for next loop
previous = current;
current = current->next;
}
// No Match
cout << "Can't remove value: no match found.\n";
}