2048游戏中的undo操作实现

Question

我已经用 C++ 实现了 2048 游戏，github 链接：2048

为了实现撤消操作，即回到之前的游戏状态，我维护了一个用于之前棋盘配置的矩阵，但如果我允许连续进行许多撤消操作，我就无法维护这么多矩阵。

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以前的移动（上、下、左或右），但如果我在这种方法中遗漏了某些东西，或者它可能是扩展，请提出一种方法来做到这一点。

Answer 1

您可以将棋盘的当前状态存储到堆栈中，因此每次用户移动时，将更改棋盘状态，只需将其放入堆栈，这样您就会得到一个堆栈，其中包含棋盘当前状态的矩阵用户的动作和从最近的一个在顶部排序。因此，当您想撤消他们的最新操作时，只需从堆栈中弹出即可，这将为您提供他们最新的操作。

...
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;
}