扫雷游戏算法

提供一个面向过程C/C++代码，很容易用类封装，改成面向对象版本的。

#include 

#include 

void makeArray(char** &pArray, int row, int col);
void printArray(char** pArray, int row,int col);
void layMines(char** pArray, int row, int col, int num);
void mineHunting(char** pArray, int row, int col, int posX, int posY);
void allMineHunting(char** pArray, int row, int col);


int main() {
    char** pArray = nullptr;
    int row = 9, col = 9, num=10;
    makeArray(pArray,row,col);
    layMines(pArray,row,col,num);
    allMineHunting(pArray,row,col);
    printArray(pArray,row,col);
    return 0;
}

void makeArray(char** &pArray, const int row, const int col){
    pArray = new char*[row];
    for(int i=0;i distribution(0,row*col);
    int count =0;
    while(count=0&&nX=0&&nY

注意，上面并不是真正的扫雷游戏算法，只是一次性将雷全部探出。真正的扫雷算法是这样的。是有一个DFS的过程在里面的。见Leetcode 529

const int direction[][2] = {{1,1},{1,0},{1,-1},{0,1},{0,-1},{-1,1},{-1,0},{-1,-1}};
class Solution {
public:

    void dfs(vector>& board, int x, int y){
        if(board[x][y]!='E') return;
        int row = board.size(), col = board[0].size();
        int count = 0;
        for(auto i:direction){
            int nx = x + i[0];
            int ny = y + i[1];
            if(nx>=0&&nx=0&&ny0){
            board[x][y] = '0' + count;
            return;
        }

        board[x][y] = 'B';
        for(auto i:direction){
            int nx = x + i[0];
            int ny = y + i[1];
            if(nx>=0&&nx=0&&ny> updateBoard(vector>& board, vector& click) {
        int x = click[0], y = click[1];
        int row = board.size(), col = board[0].size();
        if(board[x][y]=='M'){
            board[x][y] = 'X';
            return board;
        }
        dfs(board,click[0],click[1]);
        return board;
    }
};