zoukankan      html  css  js  c++  java
  • 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.

    Example 1:

    Input: 
    
    [['E', 'E', 'E', 'E', 'E'],
     ['E', 'E', 'M', 'E', 'E'],
     ['E', 'E', 'E', 'E', 'E'],
     ['E', 'E', 'E', 'E', 'E']]
    
    Click : [3,0]
    
    Output: 
    
    [['B', '1', 'E', '1', 'B'],
     ['B', '1', 'M', '1', 'B'],
     ['B', '1', '1', '1', 'B'],
     ['B', 'B', 'B', 'B', 'B']]
    
    Explanation:
    

    Example 2:

    Input: 
    
    [['B', '1', 'E', '1', 'B'],
     ['B', '1', 'M', '1', 'B'],
     ['B', '1', '1', '1', 'B'],
     ['B', 'B', 'B', 'B', 'B']]
    
    Click : [1,2]
    
    Output: 
    
    [['B', '1', 'E', '1', 'B'],
     ['B', '1', 'X', '1', 'B'],
     ['B', '1', '1', '1', 'B'],
     ['B', 'B', 'B', 'B', 'B']]
    
    Explanation:
    

    Note:

    1. The range of the input matrix's height and width is [1,50].
    2. The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
    3. The input board won't be a stage when game is over (some mines have been revealed).
    4. For simplicity, not mentioned rules should be ignored in this problem. For example, you don't need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.
     1 public class Solution {
     2     public char[][] updateBoard(char[][] board, int[] click) {
     3         int m = board.length, n = board[0].length;
     4         int x = click[0], y = click[1];
     5         
     6         if (board[x][y] == 'M') {
     7             board[x][y] = 'X';
     8         }
     9         else if (board[x][y] == 'E') {
    10             int count = countNeighbors(board, x, y);
    11             if (count == 0) {
    12                 board[x][y] = 'B';
    13                 for (int i = x - 1; i <= x + 1; i++) {
    14                     for (int j = y - 1; j <= y + 1; j++) {
    15                         if (i != x || j != y) {
    16                             if (i >= 0 && i < m && j >= 0 && j < n && board[i][j] == 'E') {
    17                                 int[] newClick = { i, j };
    18                                 updateBoard(board, newClick);
    19                             }
    20                         }
    21                     }
    22                 }
    23             }
    24             else {
    25                 board[x][y] = (char) (count + '0');
    26             }
    27         } 
    28         
    29         return board;
    30     }
    31     
    32     private int countNeighbors(char[][]board, int x, int y) {
    33         int count = 0, m = board.length, n = board[0].length;
    34         for (int i = x - 1; i < x + 2; i++) {
    35             for (int j = y - 1; j < y + 2; j++) {
    36                 if (i != x || j != y) {
    37                     if (i >= 0 && j >= 0 && i < m && j < n && board[i][j] == 'M')
    38                         count++;
    39                 }
    40             }
    41         }
    42         return count;
    43     }
    44 }
  • 相关阅读:
    .net core 部署到 iis 步骤及报错解决方法
    数据库学习笔记3 基本的查询流 2
    数据库学习笔记 2 数据库文件基本查询
    我对于C#的想法
    数据库学习笔记 一
    openwrt 软件安装依赖冲突
    openwrt 自定义DHCP
    asp.net core 3.1 入口:Program.cs中的Main函数
    c# 匿名方法(函数) 匿名委托 内置泛型委托 lamada
    家庭网络那些事
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6479949.html
Copyright © 2011-2022 走看看