529. Minesweeper（BFS）

一、题目描述

BFS在扫雷中的应用

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:

If a mine (‘M’) is revealed, then the game is over - change it to ‘X’.

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.

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.
Return the board when no more squares will be revealed.

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’],

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二，算法及代码

根据题目意思，测试的clic只可能是M或者E，而有由规则可以看出，点击M时是容易判断和返回结果的，就是直接将M换成X；而点击E时：1，如果该点E有邻居是地雷的话，需要在该位置换成邻居的地雷总数，所以找出邻居（八个方向）的地雷；2，如果该点没有邻居是地雷，则将该点换成‘B’，并且向八个方向递归地揭开，执行上述的判断方法，直到将所有的可能格子揭开。其本质就是广度优先搜索，递归逐层搜索该点周边的所有点。

class Solution {    public:    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {        if(board[click[0]][click[1]] == 'M') {            board[click[0]][click[1]] = 'X';            return board;        }        reveal(board, click[0], click[1]);        return board;    }    bool inBoard(vector<vector<char>>& board, int x, int y) {        return (x >= 0 && x < board.size() && y >= 0 && y < board[0].size());    }    void reveal(vector<vector<char>>& board,int x, int y) {        if(!inBoard(board, x, y)) return;        if(board[x][y] == 'E') {            int count = 0;            if(inBoard(board, x, y-1) && board[x][y-1] == 'M') count++;            if(inBoard(board, x-1, y-1) && board[x-1][y-1] == 'M') count++;            if(inBoard(board, x-1, y) && board[x-1][y] == 'M') count++;            if(inBoard(board, x-1, y+1) && board[x-1][y+1] == 'M') count++;            if(inBoard(board, x, y+1) && board[x][y+1] == 'M') count++;            if(inBoard(board, x+1, y+1) && board[x+1][y+1] == 'M') count++;            if(inBoard(board, x+1, y) && board[x+1][y] == 'M') count++;            if(inBoard(board, x+1, y-1) && board[x+1][y-1] == 'M') count++;            if(count > 0) {                board[x][y] = '0' + count;            } else {                board[x][y] = 'B';                reveal(board, x, y-1);                reveal(board, x-1, y-1);                reveal(board, x-1, y);                reveal(board, x-1, y+1);                reveal(board, x, y+1);                reveal(board, x+1, y+1);                reveal(board, x+1, y);                reveal(board, x+1, y-1);            }        }    }};