我必须将一个名为“reversePrint”的成员函数添加到这个双向链表中。并反向打印出名称。 例如，输入： 贾森 瑞安 迪伦 路易
输出应该是： 路易 迪伦 瑞安 杰森
我想完成这项任务的方法是首先一直向下移动到列表的末尾，即最后一个节点。然后使用“Previous”指针一次返回前一个节点，同时打印出名称。因此我会反转印刷。 但是，我认为我的指针有问题，我无法弄清楚如何修复它。一些帮助非常感谢!!!
**更新：此代码最初是单独链接的。我试图把它变成双重链接并添加“reversePrint”功能。但是我认为问题在于，我没有修改“添加”方法，所以“prev”指针是断开的。那么任何有关如何修改“添加”的帮助，以便我可以将其变成一个完全双重链接列表？
这是它现在提供的输出： prev 0x0 地址0x100200000数据a 下一个0x100300000
prev 0x0 地址0x100300000数据b 下一个0x100400000
prev 0x0 地址0x100400000数据c 下一个0x0
这是我们需要的输出： prev 0x0 地址0x100200000数据a 下一个0x100300000
prev 0x100200000 地址0x100300000数据b 下一个0x100400000
prev 0x100300000 地址0x100400000数据c 下一个0x0
以下是我的代码：
#include <iostream>
#include <string>
using namespace std;
// define a node for storage and linking
class node{
public:
string name;
node *next;
node *prev; // to be implemented by students
};
class linkedList{
public:
linkedList():top(NULL){}
bool empty(){return top == NULL;}
node *getTop(){return top;}
void setTop(node *n){top = n;}
void add(string);
int menu();
void remove(string);
~linkedList();
void reversePrint(); // to be implemented by students
friend ostream& operator << (ostream&, const linkedList&); // default output is in-order print.
private:
node *top;
node *end; // to be used for reverse print and implemented by students
};
int main(void){
linkedList l;
//cout << l.empty() << endl;
int option = 0;
string s;
bool go = true;
while(go){
option = l.menu();
switch(option){
case 1: cout << "enter a name: ";cin >> s; l.add(s); break;
case 2: cout << "enter name to be deleted: "; cin >> s; l.remove(s);break;
case 3: cout << l; break;
case 4: l.reversePrint(); break;
case 5: cout << "exiting" << endl; go = false; break;
}
}
// l goes out of scope and calls ~linkedList()
return(NULL);
}
// can not call this method "delete" - "delete" is a reserved keyword.
void linkedList::remove(string s){
bool found = false;
node *curr = getTop(), *prev=NULL;
while(curr != NULL){
// match found, delete
if(curr->name == s){
found = true;
// found at top
if(prev == NULL){
node *temp = getTop(); // delete the current node(which is the head), and set the "new head" to as the next node
setTop(curr->next);
delete(temp);
// found in list - not top
}else{
prev->next = curr->next; //Skip the current deleted node, connect previous node directly to the next node
delete(curr);
} }
// not found, advance pointers
if(!found){
prev = curr;
curr = curr->next; }
// found, exit loop
else curr = NULL; }
if(found)cout << "Deleted " << s << endl;
else cout << s << " Not Found "<< endl; }
void linkedList::add(string s){
node *n = new node();
n->name = s;
n->next = NULL;
// take care of empty list case
if(empty()){ top = n;
// take care of node belongs at beginning case
} else if(getTop()->name > s)
{
n->next = getTop();
setTop(n);
// take care of inorder and end insert
}else{
// insert in order case
node *curr = getTop(), *prev = curr;
while(curr != NULL){
if(curr->name > s)break;
prev = curr;
curr = curr->next;
}
if(curr != NULL){ // search found insert point
n->next = curr;
prev->next = n; }
// take care of end of list insertion
else if(curr == NULL){// search did not find insert point
prev->next = n; }
} }
void linkedList::reversePrint(){
node *curr = getTop(), *prev=NULL;
// jump to the last node
while(curr->next != NULL){
prev = curr;
curr = curr->next;
//curr = curr->next;
cout << "!!!!hahaha" << curr->name <<" Prev:" <<curr->prev << " " << prev->name <<endl ;//testing purpose
}
//print the name then jump back to the previous node, stops at the first node which curr->prev = NULL
while(prev != 0 ){
cout << "NULL is not 0!";
prev->prev = curr->next;
//cout << curr->name;
curr = prev;
}
}
/*ostream& operator << (ostream& os, const linkedList& ll){
//linkedList x = ll; // put this in and the code blows up - why?
node *n = ll.top;
if(n == NULL)cout << "List is empty." << endl;
else
while(n != NULL){
os << n->name << endl;
n = n->next;
} return os;
}*/
ostream& operator << (ostream& os, const linkedList& ll){
//linkedList x = ll; // put this in and the code blows up - why?
node *n = ll.top;
if(n == NULL)cout << "List is empty." << endl;
else
while(n != NULL){
os << " prev " << n->prev << endl;
os << " address " << n << " data " << n->name << endl;
os << " next " << n->next << endl;
n->prev = n;
n = n->next;
} return os;
}
// return memory to heap
linkedList::~linkedList(){
cout << "~linkedList called." << endl;
node *curr = getTop(), *del;
while(curr != NULL){
del = curr;
curr = curr->next;
delete(del);
}
}
int linkedList::menu(){
int choice = 0;
while(choice < 1 || choice > 5){
cout << "\nEnter your choice" << endl;
cout << " 1. Add a name." << endl;
cout << " 2. Delete a name." << endl;
cout << " 3. Show list." << endl;
cout << " 4. Show reverse list. " << endl; // to be implemented by students
cout << " 5. Exit. " << endl;
cin >> choice;
}
return choice;
}
答案 0 :(得分：2)
这看起来有问题：
prev->prev = curr->next;
您正在设置上一个节点的上一个节点。您可能只想设置上一个节点，对吗？
你可能想做这样的事情：
while(curr != NULL)
{
cout << curr->name;
curr = curr->prev
}
从那时起，您已确定curr是列表中的最后一个元素，您现在可以通过为其前一个元素分配curr来'递减'curr。由于第一个元素没有前一个节点，因此prev将为NULL。这样，当curr被分配给第一个元素的前一个元素时，循环将停止，该元素再次为NULL。
答案 1 :(得分：1)
您的功能添加无效。例如，考虑列表为空的情况。以下代码段将与此案例相对应。
void linkedList::add(string s){
node *n = new node();
n->name = s;
n->next = NULL;
// take care of empty list case
if(empty()){ top = n;
// take care of node belongs at beginning case
} else if(getTop()->name > s)
如您所见，未设置列表的数据成员结尾。
因此，在编写函数reversePrint的实现之前，应确保函数add正常工作。