因此,我尝试将我的java版本的链接列表移植到C++,并且使用指针指向正确的对象非常困难。删除遍历和反向遍历的目标是简单地按顺序打印链表的值,然后按相反的顺序打印。这是代码。我发布大部分内容而不是特定部分的原因是因为我相信你需要它的上下文。LinkedList C++中的指针问题
控制台输出: 它们是相等 除去5 横移 -858993460
它打印出它们是相等的,因为它立即打破在while循环的在删除(数据)的第一次迭代。
#include <iostream>
#include <string>
#include <string.h>
using namespace std;
class Node{
public:
int data;
Node *pNext;
Node();
Node(int x);
};
Node::Node(){
data = NULL;
pNext = NULL;
}
Node::Node(int x){
data = x;
pNext = NULL;
}
class List{
private:
Node *head = NULL;
public:
void insert(int data){
Node temp(data);
temp.pNext = head;
head = &temp;
}
Node removeHead(){
if (head != NULL){
Node *temp = head;
head = head->pNext;
return *temp;
}
else{
cout << "Empty List" << endl;
}
return NULL;
}
Node remove(int data){
Node *previousLink = head;
Node *currentLink = head;
while (currentLink->data != data){
if (currentLink == currentLink->pNext){
cout << "They are equal" << endl;
break;
}
previousLink = currentLink;
if (currentLink->data == NULL)
currentLink = currentLink->pNext;
cout << "Current " << currentLink->data << " Previous "
<< previousLink->data <<endl;
}
if (head->data == data){
head = head->pNext;
}
else{
previousLink->pNext = currentLink->pNext;
}
return *currentLink;
}
void traverse(){
traverse(head);
}
void reverseTraverse(){
reverseTraverse(head);
}
private:
void traverse(Node *link){
cout << link->data << endl;
if (link != link->pNext)
traverse(link->pNext);
}
void reverseTraverse(Node *link){
if (link != link->pNext)
traverse(link->pNext);
cout << link->data << endl;
}
};
int main(){
cout << " Hello World\n";
List list;
for (int i = 0; i <= 9; ++i)
list.insert(i);
list.remove(5);
cout << "removed 5\n";
cout << "traverse" << endl;
list.traverse();
cout << "reverse" << endl;
list.reverseTraverse();
system("PAUSE");
return 0;
}
它不是很清楚为什么你不使用内置的STL容器。也许使用[Standard Containers](http://www.cplusplus.com/reference/stl/)会更容易,就像'List'一样。 –
jww
2014-10-05 04:00:32