原題地址

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

題意

略

思路

簡(jiǎn)單的深度優(yōu)先，根據(jù)不同情況更新。

坑點(diǎn)

我其實(shí)沒(méi)認(rèn)真玩過(guò)掃雷擒悬。摇庙。所以做完這題才知道掃雷的規(guī)則议双，這里說(shuō)的“當(dāng)一個(gè)位置不是雷，并且周圍沒(méi)有雷的時(shí)候伊履，要遞歸地顯示出這一個(gè)位置周圍位置附近的雷的情況”，原來(lái)周圍是包括周圍8個(gè)格子的款违。唐瀑。我一開始只朝上下左右四個(gè)方向，會(huì)少更新一些值

代碼

代碼里的visited其實(shí)不需要

class Solution {
  public:

    void dfs(vector<vector<char>>& board , vector<vector<int>>&  count, vector<vector<bool>>&  visited, int x, int y) {
        int m = board.size();
        int n = board[0].size();
        if (x < 0 || y < 0 || x >= m || y >= n ) return ;
        if (visited[x][y]) return ;
        visited[x][y] = true;
        if (board[x][y] == 'M') {
            board[x][y] = 'X';
            return ;
        }
        if (board[x][y] >= '0' && board[x][y] <= '9') return ;
        if (board[x][y] == 'E') {
            if (count[x][y] == 0) {
                board[x][y] = 'B';
                dfs(board, count, visited, x - 1, y - 1);
                dfs(board, count, visited, x - 1, y);
                dfs(board, count, visited, x - 1, y + 1);
                dfs(board, count, visited, x, y - 1);
                dfs(board, count, visited, x, y + 1);
                dfs(board, count, visited, x + 1, y - 1);
                dfs(board, count, visited, x + 1, y );
                dfs(board, count, visited, x + 1, y + 1);
                // dfs(board, count, visited, x + 1, y);
                // dfs(board, count, visited, x - 1, y);
                // dfs(board, count, visited, x, y + 1);
                // dfs(board, count, visited, x, y - 1);
            } else {
                board[x][y] = (char)(count[x][y] + '0');
            }
        }
    }

    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        if (click.empty()) return board;
        vector<vector<char>> result;
        if (board.empty()) return result;
        int m = board.size();
        int n = board[0].size();
        vector<vector<int>> count(m, vector<int>(n, 0));
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        countMinesAround(board, count);
        dfs(board, count, visited, click[0], click[1]);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                cout << count[i][j] << " ";
            }
            cout << endl;
        }
        return board;
    }

    void countMinesAround(vector<vector<char>>& board, vector<vector<int>>&  count) {
        int m = board.size();
        int n = board[0].size();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i - 1 >= 0 && j - 1 >= 0) {
                    count[i][j] += (board[i - 1][j - 1] == 'M' ? 1 : 0);
                }
                if (i - 1 >= 0) {
                    count[i][j] += (board[i - 1 ][j ] == 'M' ? 1 : 0);
                }
                if (i - 1 >= 0 && j + 1 < n) {
                    count[i][j] += (board[i - 1][j + 1] == 'M' ? 1 : 0);
                }
                if (j - 1 >= 0) {
                    count[i][j] += (board[i ][j - 1] == 'M' ? 1 : 0);
                }
                if (j + 1 >= 0) {
                    count[i][j] += (board[i ][j + 1] == 'M' ? 1 : 0);
                }
                if (i + 1 < m && j - 1 >= 0) {
                    count[i][j] += (board[i + 1][j - 1] == 'M' ? 1 : 0);
                }
                if (i + 1 < m) {
                    count[i][j] += (board[i + 1][j] == 'M' ? 1 : 0);
                }
                if (i + 1 < m && j + 1 < n) {
                    count[i][j] += (board[i + 1][j + 1] == 'M' ? 1 : 0);
                }

            }
        }
    }
};