地雷游戏 Minesweeper 原

Let's play the minesweeper game (Wikipediaonline 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】给定一个地图和一系列的点击位置，需要根据点击的情况对地图进行更新。地图中：

M表示未探明的地雷,
E表示未探明的安全区,
B表示已探明且周围8个格子均没有地雷的安全区,
X表示点击中了地雷,

1. 如果点击地雷（'M'），将其标记为“X”，停止进一步搜索。
2. 如果点击一个未探明的安全区（'E'），取决于多少个周围的地雷：

2.1 周围有地雷，标上周围地雷的数量，停止进一步的搜索。

2.2 周围没有有地雷，标记为“B”，继续搜索其邻居。

①  对于当前需要点击的点，我们先判断是不是雷，是的话直接标记X返回即可。如果不是的话，我们就数该点周围的雷个数，如果周围有雷，则当前点变为雷的个数并返回。如果没有的话，我们再对周围所有的点调用递归函数（dfs）再点击即可。

class Solution { //8ms
public char[][] updateBoard(char[][] board, int[] click) {
int m = board.length;
int n = board.length;
int row = click;
int col = click;
int count = 0;
if (board[row][col] == 'M'){//点到地雷
board[row][col] = 'X';
}else {//未标明的安全区域
for (int i = -1;i < 2;i ++){//计算周围地雷的数量
for (int j = -1;j < 2;j ++){
if (i == 0 && j == 0) continue;
int x = row + i;
int y = col + j;
if (x < 0 || x >= m || y < 0 || y >= n) continue;
if (board[x][y] == 'M' || board[x][y] == 'X') count ++;
}
}
if (count > 0){//如果周围有地雷，返回其周围的地雷数
board[row][col] = (char) (count + '0');
}else {//标记为B，继续遍历周围的点
board[row][col] = 'B';
for (int i = -1;i < 2;i ++){
for (int j = -1;j < 2;j ++){
if (i == 0 && j == 0) continue;
int x = row + i;
int y = col + j;
if (x < 0 || x >= m || y < 0 || y >= n) continue;
if (board[x][y] == 'E'){//未探明的安全区域
updateBoard(board,new int[]{x,y});//递归向周围遍历
}
}
}
}
}
return board;
}
}

② BFS。

class Solution { //12ms
public char[][] updateBoard(char[][] board, int[] click) {
int m = board.length;
int n = board.length;
queue.offer(click);
while (! queue.isEmpty()){
int[] cell = queue.poll();
int row = cell;
int col = cell;
if (board[row][col] == 'M'){
board[row][col] = 'X';
}else {
int count = 0;
for (int i = -1;i < 2;i ++){
for (int j = -1;j < 2;j ++){
if (i == 0 && j == 0) continue;
int x = row + i;
int y = col + j;
if (x < 0 || x >= m || y < 0 || y >= n) continue;
if (board[x][y] == 'M' || board[x][y] == 'X') count ++;
}
}
if (count > 0){
board[row][col] = (char)(count + '0');
}else {
board[row][col] = 'B';
for (int i = -1;i < 2;i ++){
for (int j = -1;j < 2;j ++){
if (i == 0 && j == 0) continue;
int x = row + i;
int y = col + j;
if (x < 0 || x >= m || y < 0 || y >= n) continue;
if (board[x][y] == 'E'){
queue.offer(new int[]{x,y});
board[x][y] = 'B';
}
}
}
}
}
}
return board;
}
}

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