【LeetCode】529. Minesweeper

问题描述

问题链接：

Let’s play the minesweeper game (Wikipedia, online game)!

You are given a 2D char matrix representing the game board. ‘M’ represents an unrevealed mine, ‘E’ represents an unrevealed empty square, ‘B’ represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit (‘1’ to ‘8’) represents how many mines are adjacent to this revealed square, and finally ‘X’ represents a revealed mine.

Now given the next click position (row and column indices) among all the unrevealed squares (‘M’ or ‘E’), return the board after revealing this position according to the following rules:

Example 1:

Input: 

[['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'M', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E']]

Click : [3,0]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Note:

1. The range of the input matrix’s height and width is [1,50].
2. The click position will only be an unrevealed square (‘M’ or ‘E’), which also means the input board contains at least one clickable square.
3. The input board won’t be a stage when game is over (some mines have been revealed).
4. For simplicity, not mentioned rules should be ignored in this problem. For example, you don’t need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.

我的代码

public class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        /*
        思路，主要难点是处理B，首先看自己周边有没有雷，有的话把自己变成数字，没有的话就递归的处理自己周围的空格
        */
        int x = click[0];
        int y = click[1];
        if(board[x][y] == 'M'){
            dealM(board,click);
        }else if(board[x][y] == 'E'){
            dealE(board,click[0],click[1]);
        }
        return board;
    }

    private void dealM(char[][] board, int[] click){
        int x = click[0];
        int y = click[1];
        board[x][y] = 'X';
    }

    private void dealE(char[][] board, int x, int y){

        int m = board.length;
        int n = board[0].length;

        int count = getMineCount(board,x,y); // 临近雷的数量

        if(count > 0){
            board[x][y] = (char)(count + (int)('0'));
        }else{
            board[x][y] = 'B';
            doRecursive(board,x,y);
        }


    }

    private void doRecursive(char[][] board,int x, int y){
        int m = board.length;
        int n = board[0].length;
        // 左
        if(y > 0 && board[x][y - 1] == 'E'){

            dealE(board,x,y-1);
        }
        // 右
        if(y < n - 1 && board[x][y + 1] == 'E'){
            dealE(board,x,y+1);
        }
        // 上
        if(x > 0 && board[x - 1][y] == 'E'){
            dealE(board,x-1,y);
        }
        // 下
        if(x < m - 1 && board[x + 1][y] == 'E'){
            dealE(board,x+1,y);
        }
        // 左上
        if(y > 0 && x > 0 && board[x - 1][y - 1] == 'E'){
            dealE(board,x-1,y-1);
        }
        // 右上
        if(x > 0 && y < n - 1 && board[x - 1][y + 1] == 'E'){
            dealE(board,x-1,y+1);
        }
        // 左下
        if(x < m - 1 && y > 0 && board[x + 1][y - 1] == 'E'){
            dealE(board,x+1,y-1);
        }
        // 右下
        if(y < n - 1 && x < m - 1 && board[x + 1][y + 1] == 'E'){
            dealE(board,x+1,y+1);
        }

    }

    private int getMineCount(char[][] board, int x, int y){
        int m = board.length;
        int n = board[0].length;
        int count = 0;
        // 左
        if(y > 0 && board[x][y - 1] == 'M'){
            count++;
        }
        // 右
        if(y < n - 1 && board[x][y + 1] == 'M'){
            count++;
        }
        // 上
        if(x > 0 && board[x - 1][y] == 'M'){
            count++;
        }
        // 下
        if(x < m - 1 && board[x + 1][y] == 'M'){
            count++;
        }
        // 左上
        if(y > 0 && x > 0 && board[x - 1][y - 1] == 'M'){
            count++;
        }
        // 右上
        if(x > 0 && y < n - 1 && board[x - 1][y + 1] == 'M'){
            count++;
        }
        // 左下
        if(x < m - 1 && y > 0 && board[x + 1][y - 1] == 'M'){
            count++;
        }
        // 右下
        if(y < n - 1 && x < m - 1 && board[x + 1][y + 1] == 'M'){
            count++;
        }
        return count;
    }
}

打败了70.30%的Java代码。来看看讨论区。

讨论区

基本跟我的思路差不多，就不放了。