undo操作在2048游戏中的实现

Question

我已经在C++中实现了2048游戏，github链接：2048

为了实现撤销操作，也就是返回到之前的游戏状态，我为之前的棋盘配置维护了一个矩阵，但是如果我允许许多连续的撤销操作，我就不能保持矩阵的数量。

有什么方法可以改进这种方法？

我认为的一种方法是只保留之前的移动(向上、向下、向左或向右)，但只有这些信息不能帮助重新生成以前的状态，如果我在这种方法中遗漏了什么，或者它可以扩展，请建议一种方法来做到这一点。

Answer 1

你可以将棋盘的当前状态存储到堆栈中，所以每次用户进行移动时，只要将其放入堆栈中，就可以得到一个堆栈，其中充满了用户移动的棋盘当前状态矩阵，并从最新的一个开始排序。因此，当你想要撤销他们的最新动作时，只需从堆栈中弹出，这将为你提供他们的最新操作。

...
std::stack<std::array<int, 16>> boardStates;
std::array<16, int> currentBoardState;
// whenever user makes a move
makeMove(currentBoardState)

//when they want to undo    
tempBoardState = undoMove();
if(tempBoardState != nullptr)
{
   currentBoardState = tempBoardState;
}
else
{
   std::cout << "No previous move available" << std::endl
}
...

void makeMove(std::array<int, 16> currentState)
{
   boardStates.push(currentState);
}

std::array<int, 16> undoMove()
{
    if(!boardStates.empty())
    {
        return boardStates.pop();
    }

    return nullptr;
}

Answer 2

实现历史记录，以存储后续的板状态更改。

//maximum history size (can be whatever reasonable value)
const int MAX_UNDO = 2048;

//give a shorter name to the type used to store the board status
typedef std::array<int, 16> snapshot; 

//use a double ended queue to store the board statuses
std::deque<snapshot> history;

//this function saves the current status, has to be called each time 
//the board status changes, i.e. after the board initialization 
//and after every player move
void save()
{
    //make a copy of g, the current status
    snapshot f;
    std::copy(&g[0][0], &g[0][0] + 16, f.begin());

    //push the copy on top
    history.push_back(f); 

    //check history size
    if(history.size() > MAX_UNDO)
    {
        //remove one element at the bottom end
        history.pop_front();
    }
}

bool undo()
{
    //history must hold at least one element 
    //other than the current status copy
    if(history.size() > 1)
    {
        //the top element of the queue always holds a copy of the 
        //current board status: remove it first
        history.pop_back(); 

        //now the top element is the previous status copy
        snapshot f = history.back();

        //copy it back to g
        std::copy(f.begin(), f.end(), &g[0][0]);

        //undo operation succedeed
        return true;
    }

    //undo operation failed
    return false;
}