zoukankan      html  css  js  c++  java
  • LC 529. Minesweeper

    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.

    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;
      }
    };
  • 相关阅读:
    猜数字游戏(补)
    团队项目五(项目回顾)
    项目回顾
    第二次阶段冲刺
    团队项目(任务三):第一次冲刺
    个人项目(一):新猜数字
    课后作业(一)
    团队任务二
    团队任务(一)
    课后作业(一)
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10255316.html
Copyright © 2011-2022 走看看