提供一個面向過程C/C++代碼，很容易用類封裝，改成面向對象版本的。

#include <iostream>#include <random>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<col;i++){pArray[i] = new char[col];}for(int i=0;i<row;i++){for(int j=0;j<col;j++){pArray[i][j] = '0';}} }void printArray(char** pArray,int row,int col){for(int i=0;i<row;i++){for(int j=0;j<col;j++){std::cout<<pArray[i][j]<<" ";}std::cout<<std::endl;} }void layMines(char** pArray,int row,int col, int num){std::random_device generator;std::uniform_int_distribution<int> distribution(0,row*col);int count =0;while(count<num){int dice_roll = distribution(generator);int i = dice_roll/row;int j = dice_roll%col;if(pArray[i][j]!='9'){pArray[i][j]='9';count++;}}}void mineHunting(char** pArray, int row, int col, int posX, int posY){if(pArray[posX][posY]!='0') return;int count = 0;const int direction[][2] = {{1,1},{1,0},{1,-1},{0,1},{0,-1},{-1,1},{-1,0},{-1,-1}};for(auto i : direction){int nX = posX + i[0];int nY = posY + i[1];if(nX>=0&&nX<row&&nY>=0&&nY<col&&pArray[nX][nY]=='9') count++;}pArray[posX][posY] = char('0' + count);}void allMineHunting(char** pArray, int row, int col){for(int i=0;i<row;i++){for(int j=0;j<col;j++){mineHunting(pArray,row,col,i,j);}} }

注意，上面并不是真正的掃雷游戲算法，只是一次性將雷全部探出。真正的掃雷算法是這樣的。是有一個DFS的過程在里面的。見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<vector<char>>& 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<row&&ny>=0&&ny<col&&board[nx][ny]=='M') count++;}if(count>0){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<row&&ny>=0&&ny<col){dfs(board,nx,ny);}}}vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& 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;} };

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