其实就是找连通分量的问题，使用 DFS 或者 BFS 均可。
方法一：深度优先遍历
- M 地雷
- E 还未挖出的空方块
- B 已经挖出的空方块
public class Solution {
/**
* 相邻关系规定为：8 个方向
*/
private int[][] directions = {
{
0, 1}, {
1, 0}, {
0, -1}, {
-1, 0}, {
1, 1}, {
-1, -1}, {
1, -1}, {
-1, 1}};
private int rows;
private int cols;
public char[][] updateBoard(char[][] board, int[] click) {
this.rows = board.length;
this.cols = board[0].length;
int x = click[0];
int y = click[1];
if (board[x][y] == 'M') {
// 规则 1
board[x][y] = 'X';
return board;
}
dfs(board, x, y);
return board;
}
public void dfs(char[][] board, int x, int y) {
// 相邻地雷的数量
int count = 0;
for (int[] direction : directions) {
int newX = x + direction[0];
int newY = y + direction[1];
if (inArea(newX, newY) && board[newX][newY] == 'M') {
count++;
}
}
if (count > 0) {
// 规则 3
board[x][y] = (char) (count + '0');
} else {
// 规则 2：如果当前位置没有地雷，将它修改为 B
board[x][y] = 'B';
for (int[] direction : directions) {
int newX = x + direction[0];
int newY = y + direction[1];
if (inArea(newX, newY) && board[newX][newY] == 'E') {
dfs(board, newX, newY);
}
}
}
}
private boolean inArea(int x, int y) {
return x >= 0 && x < rows && y >= 0 && y < cols;
}
}
方法二：广度优先遍历
import java.util.LinkedList;
import java.util.Queue;
public class Solution {
private int[][] directions = {
{
0, 1}, {
1, 0}, {
0, -1}, {
-1, 0}, {
1, 1}, {
-1, -1}, {
1, -1}, {
-1, 1}};
private int rows;
private int cols;
public char[][] updateBoard(char[][] board, int[] click) {
int x = click[0];
int y = click[1];
if (board[x][y] == 'M') {
board[x][y] = 'X';
return board;
}
this.rows = board.length;
this.cols = board[0].length;
boolean[][] visited = new boolean[rows][cols];
visited[x][y] = true;
Queue<int[]> queue = new LinkedList<>();
queue.offer(new int[] {
x, y});
while (!queue.isEmpty()) {
int[] point = queue.poll();
int i = point[0];
int j = point[1];
int count = 0;
for (int[] direction:directions) {
int newX = i + direction[0];
int newY = j + direction[1];
if (inArea(newX,newY) &&board[newX][newY] == 'M') {
count++;
}
}
if (count > 0) {
board[i][j] = (char)(count + '0');
} else {
board[i][j] = 'B';
for (int[] direction:directions) {
int newX = i + direction[0];
int newY = j + direction[1];
if (inArea(newX,newY) && !visited[newX][newY] && board[newX][newY] == 'E') {
visited[newX][newY] = true;
queue.offer(new int[] {
newX, newY});
}
}
}
}
return board;
}
private boolean inArea(int x, int y) {
return x >= 0 && x < rows && y >= 0 && y < cols;
}
}