(dfs) leetcode 528. Minesweeper

思路：这道题和小岛那道有些类似，主要是实现以下这四个操作：

1. If a mine ('M') is revealed, then the game is over - change it to 'X'.   若点到M，它是代表还未发现的炸弹，则该位置变为X，游戏结束。
2. If an empty square ('E') with no adjacent mines is revealed, then change it to revealed blank ('B') and all of its adjacent unrevealed squares should be revealed recursively.    如果点到E，它的四周没有炸弹，则该位置变为B，然后递归遍历它附近的8个格子（用dfs）。
3. If an empty square ('E') with at least one adjacent mine is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.   如果点到E，它的四周至少有一个炸弹，则把该位置变为1-8（为炸弹的个数）。
4. Return the board when no more squares will be revealed.   返回board。

class Solution {
public:
    int dx[8] = {-1,-1,-1, 1, 1, 1,0,0};
    int dy[8] = {-1, 0, 1,-1, 0, 1,-1,1};
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        //If a mine ('M') is revealed, then the game is over - change it to 'X'.
        int x = click[0], y = click[1];
        if(board[x][y]=='M'){
            board[x][y] = 'X';
            return board;
        }
        //If an empty square ('E') with no adjacent mines is revealed, then change it to revealed blank ('B') and all of its adjacent unrevealed squares should be revealed recursively.
        dfs(board, x, y);
        return board;
        
    }
    
    void dfs(vector<vector<char>>& board, int x, int y){
        int m = board.size(), n = board[0].size();
        if(x<0 || x>=m || y<0 || y>=n || board[x][y] !='E')
            //出口
            return;
        int num = numOfBombs(board, x, y);
        //1. no adjacenet mine
        if(num==0){
            board[x][y] = 'B';
            for(int i=0; i<8; i++){
                int newX = x + dx[i], newY = y + dy[i];
                dfs(board, newX, newY);
            }
        }
        else{
            //2. at least one mine
            board[x][y] = num +'0';
        }
    }
    
    int numOfBombs(vector<vector<char>>& board, int x, int y){
        int num = 0;  //记录附近8个格子中炸弹的个数
        int m = board.size(), n = board[0].size();
        for(int i=0; i<8; ++i){
            //遍历附近的8格
            int newX = x +dx[i], newY = y +dy[i];
            if(newX<0 || newX>=m || newY<0 || newY>=n)
                continue;
            if(board[newX][newY]=='M' || board[newX][newY]=='X')
                ++num;
        }
        return num;
    }
};

 

转载于:https://www.cnblogs.com/Bella2017/p/11165914.html