LeetCode:529. Minesweeper

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:

1. If a mine (‘M’) is revealed, then the game is over - change it to ‘X’.
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.
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.
4. 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'], ['B', 'B', 'B', 'B', 'B']]Explanation:

Example 2:

Input: [['B', '1', 'E', '1', 'B'], ['B', '1', 'M', '1', 'B'], ['B', '1', '1', '1', 'B'], ['B', 'B', 'B', 'B', 'B']]Click : [1,2]Output: [['B', '1', 'E', '1', 'B'], ['B', '1', 'X', '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.

题意：编写扫雷的action程序，重点注意例子1中的(0,2)并没有变化，也就是在宽度搜索时，它周围的节点都没有加入队列，而之所以会这样，是因为它的下方有雷。换句话说，只要某个位置周围有雷，就不加入搜索队列。所以必须先判断当前位置存在雷的情况，然后再决定是否加入搜索队列中。
代码如下：

class Solution {    int[][]actions={{0,1},{0,-1},{1,0},{-1,0},{1,1},{-1,-1},{1,-1},{-1,1}};    public char[][] updateBoard(char[][] board, int[] click) {        int n=board.length;        int m=board[0].length;        LinkedList<Integer> x=new LinkedList<Integer>();        LinkedList<Integer> y=new LinkedList<Integer>();        if(board[click[0]][click[1]]=='M')        {                board[click[0]][click[1]]='X';            return board;        }        board[click[0]][click[1]]=getCount(board,n,m,click[0],click[1]);//首先判断click位置的存雷情况        if(board[click[0]][click[1]]!='B')//click位置周围有雷，只变化click位置            return board;        x.addFirst(click[0]);        y.addFirst(click[1]);        while(x.size()>0){            int xt=x.removeLast();            int yt=y.removeLast();            for(int i=0;i<8;i++){                int tx=xt+actions[i][0];                int ty=yt+actions[i][1];                if(tx>=0&&tx<n&&ty>=0&&ty<m&&board[tx][ty]=='E'){                    board[tx][ty]=getCount(board,n,m,tx,ty);                    if(board[tx][ty]=='B')                    {                        x.addFirst(tx);                        y.addFirst(ty);                    }                }            }        }        return board;    }    public char getCount(char[][]board,int n,int m,int xt,int yt){//计算四周存在雷的个数        int count=0;        for(int i=0;i<8;i++){            int tx=xt+actions[i][0];            int ty=yt+actions[i][1];            if(tx>=0&&tx<n&&ty>=0&&ty<m){                if(board[tx][ty]=='M'){                    count++;                }            }         }        return (char) (count>0?'0'+count:'B');    }}