LeetCode 529 Minesweeper (BFS)

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.

题目分析：主要需要处理扩展情况，按照BFS遍历即可，对于每个点判断是否是可扩展点，不是的话计算雷的个数 (9ms 击败44%)

public class Solution {    public static final int[] dirx = {1, 1, 1, -1, -1, -1, 0, 0};    public static final int[] diry = {1, -1, 0, 1, -1, 0, 1, -1};    public static boolean[][] vis;    public static int n, m;        public static class POINT {        int x, y;        POINT(int x, int y) {            this.x = x;            this.y = y;        }    }    public static int calSurround(int x, int y, char[][] board) {          int cnt = 0;        for (int i = 0; i < 8; i ++) {            int xx = x + dirx[i];            int yy = y + diry[i];            if (xx >= 0 && yy >= 0 && xx < n && yy < m && board[xx][yy] == 'M') {                cnt ++;            }        }        return cnt;    }    public static void BFS(int x, int y, char[][] board) {        Queue<POINT> q = new LinkedList<>();        q.add(new POINT(x, y));        vis[x][y] = true;        while (!q.isEmpty()) {            POINT cur = q.poll();            if (board[cur.x][cur.y] == 'E') {                board[cur.x][cur.y] = 'B';            }            int cnt = calSurround(cur.x, cur.y, board);            if (cnt == 0) {                for (int i = 0; i < 8; i ++) {                    int xx = cur.x + dirx[i];                    int yy = cur.y + diry[i];                    if (xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy]) {                        vis[xx][yy] = true;                        q.add(new POINT(xx, yy));                    }                }            } else {                board[cur.x][cur.y] = (char) ('0' + cnt);            }        }    }    public char[][] updateBoard(char[][] board, int[] click) {        n = board.length;        m = board[0].length;        vis = new boolean[n + 1][m + 1];        int x = click[0];        int y = click[1];        if (board[x][y] == 'M') {            board[x][y] = 'X';            return board;        }        for (int i = 0; i <= n; i ++) {            Arrays.fill(vis[i], false);        }        BFS(x, y, board);        return board;    }}