LIST變形 C++

假設現在有兩個LIST(圖片one two)，我可以選擇要動哪一個list(選擇動哪張圖片移動到新的位置)，如果沒動到的那張圖片，希望也要出一個空間，存的XY就是原本的位置。
此時希望可以有恢復的功能(linking list的返回鍵left)，也希望可以下一步(linking list的right)

#include<iostream>
using namespace std;
struct list
{
list* left;
int x;
int y;
list* right;
};
int main()
{
list one;
list two;
cout<<"請輸入第1張圖的X和Y位置"<<endl;
cin>>one.x;
cin>>one.y;
one.left=NULL;
one.right=new list;
cout<<"請輸入第2張圖的X和Y位置"<<endl;
cin>>two.x;
cin>>two.y;
two.left=NULL;
two.right=new list;
list* pre;
list* current;
list* next;
int choice,count=0;
while(true)
{
cout<<"請輸入要更動哪一個圖片one or two"<<endl;
cin>>choice;
if(choice==1)
{
current=one.right; //initial 1
pre=&one;
for(int i=0;i<count;i++) //find new one
{
pre=current;
next=current->right;
current=next;
}
cout<<"請輸入第1張圖新的X和Y位置"<<endl;
cin>>current->x; //new
cin>>current->y;
current->left=pre;
current->right=new list;
current=two.right; //initial 2
pre=&two;
for(int j=0;j<count;j++) //find new two
{
pre=current;
next=current->right;
current=next;
}
current->left=pre;
current->right=new list;
current->x=current->left->x; //copy two data
current->y=current->left->y;
count++;
}
else if(choice==2)
{
current=two.right; //initial 2
pre=&two;
for(int j=0;j<count;j++) //find new two
{
pre=current;
next=current->right;
current=next;
}
cout<<"請輸入第2張圖新的X和Y位置"<<endl;
cin>>current->x; //new
cin>>current->y;
current->left=pre;
current->right=new list;
current=one.right; //initial 1
pre=&one;
for(int i=0;i<count;i++) //find new one
{
pre=current;
next=current->right;
current=next;
}
current->left=pre;
current->right=new list;
current->x=current->left->x; //copy one data
current->y=current->left->y;
count++;
}
else
break;
}

/*trace 使用
current=&one;
for(int k=0;k<=count;k++)
{
cout<<current->x<<"   "<<current->y<<endl;
next=current->right;
current=next;
}
*/
return 0;
}

當然此程式可以看出choice==one or two 內有很多地方是做一樣的事情，應該可以額外寫個function去genrealize，但是由於懶惰，就不想優化了....