• 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:
    xml读写
    scrollWidth,clientWidth与offsetWidth的区别
    DIV+CSS设计时浏览器兼容性
    访问IIS客户端出现server application error解决方案
    网站局域网内不能访问解决方法之一
    xml学习笔记(一)
    文本编辑器FCKeditor
    业务部门需要IT人员为其提供什么
    程序员与VBA之怪现状
    你的代码完成了吗?(之一)——可用性和易用性
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10255316.html
Copyright © 2020-2023  润新知