Minesweeper

https://leetcode.com/problems/minesweeper/description/

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:

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.

Thoughts

这道题的意思是如果碰到雷就立马结束。否则继续遍历，周围有雷，就把B更新为雷的数目，并不再对它后面的进行遍历， M也是。只有周围没雷的B才继续往后遍历。

Code

class Solution {
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        queue<pair<int, int>> q;
        auto x = click[0], y = click[1];
        if (board[x][y] == 'M') {
            board[x][y] = 'X';
        } else {
            q.emplace(x, y);
        }

        const auto m = board.size(), n = board[0].size();
        while (!q.empty()) {
            const auto size = q.size();
            for (int i = 0; i < size; ++i) {
                const auto p = q.front(); q.pop();
                int cur_x = p.first, cur_y = p.second;
                if (board[cur_x][cur_y] != 'E') continue;
                int count = 0;
                vector<pair<int, int>> tmp;
                for (int j = -1; j <= 1; ++j) {
                    for (int k = -1; k <= 1; ++k) {
                        x = cur_x + j;
                        y = cur_y + k;
                        if (x >= 0 && x < m && y >= 0 && y < n) {
                            if (board[x][y] == 'M') ++count;
                            tmp.emplace_back(x, y);
                        }
                    }
                }
                if (count) {
                    board[cur_x][cur_y] = '0' + count;
                } else {
                    board[cur_x][cur_y] = 'B';
                    for (const auto p : tmp) {
                        q.push(p);
                    }
                }
            }
        }
        return board;
    }
};

class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        int[][] dirs = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {-1, 1}, {-1, -1}, {1, -1}, {1, 1}};
        int m = board.length, n = board[0].length;

        if (board[click[0]][click[1]] == 'M') {
            board[click[0]][click[1]] = 'X';
            return board;
        }

        Queue<int[]> queue = new LinkedList<>();
        queue.add(click);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] pos = queue.poll();
                char c = board[pos[0]][pos[1]];
                if (c == 'M' || c == 'B') {
                    continue;
                }

                int count = 0;
                for (int[] dir : dirs) {
                    int x = pos[0] + dir[0];
                    int y = pos[1] + dir[1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        char b = board[x][y];
                        if (b == 'M') {
                            count++;
                        }
                    }
                }

                if (count != 0) {
                    board[pos[0]][pos[1]] = (char)(count + '0');
                    continue;
                }

                board[pos[0]][pos[1]] = 'B';
                for (int[] dir : dirs) {
                    int x = pos[0] + dir[0];
                    int y = pos[1] + dir[1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        queue.add(new int[]{x, y});
                    }
                }
            }
        }

        return board;
    }
}

Analysis

Errors:

1. 对于当前点, 没判断之前是否遍历过了

时空复杂度都是O(MN)