标签:fas change oar wiki lse when least play number
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:
Runtime: 28 ms, faster than 84.38% of C++ online submissions for Minesweeper.
class Solution { public: vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) { deque<pair<int,int>> q({ {click[0], click[1]} }); while(!q.empty()){ auto c = q.front().first, r = q.front().second, mines = 0; vector<pair<int,int>> neighbours; if(board[c][r] == ‘M‘) board[c][r] = ‘X‘; else for(int i=-1; i<=1; i++){ for(int j=-1; j<=1; j++){ if(c+i >= 0 && r+j >= 0 && c+i < board.size() && r+j < board[0].size()){ if(board[c+i][r+j] == ‘M‘) ++mines; else if(mines == 0 && board[c+i][r+j] == ‘E‘) neighbours.push_back({c+i,r+j}); } } } if(mines > 0) board[c][r] = ‘0‘ + mines; else for(auto n : neighbours) { board[n.first][n.second] = ‘B‘; q.push_back(n); } q.pop_front(); } return board; } };
标签:fas change oar wiki lse when least play number
原文地址:https://www.cnblogs.com/ethanhong/p/10255316.html