-2
我必須把這個名爲「reversePrint」的成員函數添加到這個雙向鏈表中。並反向打印出名字。 例如，輸入： 傑森 瑞安 迪倫 路易雙向鏈接列表反向打印？
輸出應該是： 路易 迪倫 瑞安 傑森
我能完成這一操作思路是向下移動一直到結束列表中的第一個，即最後一個節點。然後使用「Previous」指針一次返回到前一個節點，同時打印出名稱。所以我會有一個反轉的印刷。 但是，我的指針有點問題，我想不出如何解決它。一些幫助真的很感謝！
**更新：此代碼最初是單鏈接的。我試圖把它變成雙重鏈接並添加「reversePrint」功能。不過我認爲問題在於，我沒有修改「添加」方法，因此「prev」指針斷開。那麼，如何修改「添加」的任何幫助，使我可以成爲一個完全雙向鏈表？
這是它現在給出的輸出： 分組爲0x0 地址0x100200000數據的 下一0x100300000
分組爲0x0 地址0x100300000數據B 下0x100400000
分組爲0x0 地址0x100400000數據c next 0x0
這是我們需要的輸出： prev 0x0 個地址0x100200000數據 下一0x100300000
分組0x100200000 地址0x100300000數據B 未來0x100400000
分組0x100300000 地址0x100400000數據c 未來爲0x0
以下是我的代碼：
#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;
}
這怎麼可能是你的代碼？ – 2013Asker
這就像「正向打印」。只有從「尾巴」（vs.「head」）開始，並通過「previousNode」（與「nextNode」）進行導航。如果這不起作用，則在創建列表時可能沒有正確設置前一個節點指針。 – user2864740
這是個玩笑嗎？ –