529. Minesweeper

Posted 2020-10-01 apanda009

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\'],
 [\'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:
The range of the input matrix\'s height and width is [1,50].
The click position will only be an unrevealed square (\'M\' or \'E\'), 
which also means the input board contains at least one clickable square.
The input board won\'t be a stage when game is over (some mines have been revealed).
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.

 

BFS

1. matrix 邻居都是找好的, 但是此题是八邻居问题

2. visited  数组是防止邻居重叠遍历的问题和重复加入的问题, 而此题需要遍历所有的邻居 但是此题再往队列加的时候判断是否为\'E\', 并且将其转化为\'B\'

3. and all of its adjacent unrevealed squares should be revealed recursively. 当邻居为\' E \'时, 需要加入队了, 防止加重复, 更新为 B

如果此处不更新邻居 或不判断就加的话将造成内存溢出, 重复加入

// if (board[r][c] == \'E\') {
queue.add(new int[] {r, c});
board[r][c] = \'B\'; // Avoid to be added again.
// }

public char[][] updateBoard(char[][] board, int[] click) {
        int m = board.length, n = board[0].length;
        Queue<int[]> queue = new LinkedList<>();
        queue.add(click);
        
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0], col = cell[1];
            
            if (board[row][col] == \'M\') { // Mine
                board[row][col] = \'X\';
            }
            else { // Empty
                // Get number of mines first.
                int count = 0;
                for (int i = -1; i < 2; i++) {
                    for (int j = -1; j < 2; j++) {
                        if (i == 0 && j == 0) continue;
                        int r = row + i, c = col + j;
                        if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                        if (board[r][c] == \'M\' || board[r][c] == \'X\') count++;
                    }
                }
                
                if (count > 0) { // If it is not a \'B\', stop further DFS.
                    board[row][col] = (char)(count + \'0\');
                }
                else { // Continue BFS to adjacent cells.
                    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 r = row + i, c = col + j;
                            if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                            if (board[r][c] == \'E\') {
                                queue.add(new int[] {r, c});
                                board[r][c] = \'B\'; // Avoid to be added again.
                            }
                        }
                    }
                }
            }
        }
        
        return board;
    }

DFS

不用队,但是邻居是调用自己这个函数的, 需要遍历所有的邻居, 并且dfs时已经改了路上的E, 因此不需要visited 和转B

if (board[r][c] == \'E\') updateBoard(board, new int[] {r, c});

public char[][] updateBoard(char[][] board, int[] click) {
        int m = board.length, n = board[0].length;
        int row = click[0], col = click[1];
        
        if (board[row][col] == \'M\') { // Mine
            board[row][col] = \'X\';
        }
        else { // Empty
            // Get number of mines first.
            int count = 0;
            for (int i = -1; i < 2; i++) {
                for (int j = -1; j < 2; j++) {
                    if (i == 0 && j == 0) continue;
                    int r = row + i, c = col + j;
                    if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                    if (board[r][c] == \'M\' || board[r][c] == \'X\') count++;
                }
            }
            
            if (count > 0) { // If it is not a \'B\', stop further DFS.
                board[row][col] = (char)(count + \'0\');
            }
            else { // Continue DFS to adjacent cells.
                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 r = row + i, c = col + j;
                        if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                        if (board[r][c] == \'E\') updateBoard(board, new int[] {r, c});
                    }
                }
            }
        }
        
        return board;
    }