[LeetCode] 529. Minesweeper 扫雷
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:
- If a mine ('M') is revealed, then the game is over - change it to 'X'.
- 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.
- 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.
- 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:
- The range of the input matrix's height and width is [1,50].
- The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
- The input board won't be a stage when game is over (some mines have been revealed).
- 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.
扫雷游戏，经典的搜索问题。
1. 当点击到雷('M')时，标记为'X'，结束搜索，游戏结束。
2. 当点击到空方块('E')时，如果周围有雷，就计算雷的个数，标记为这个数字，不在搜索。如果周围没有雷的话， 标记为'B'，继续搜索8个挨着的方块。
解法1：DFS
解法2： BFS
Java: DFS
public class Solution {
public char[][] updateBoard(char[][] board, int[] click) {
int m = board.length, n = board[0].length;
int row = click[0], col = click[1];
if (board[row][col] == 'M') { // Mine
board[row][col] = 'X';
}
else { // Empty
// Get number of mines first.
int count = 0;
for (int i = -1; i < 2; i++) {
for (int j = -1; j < 2; j++) {
if (i == 0 && j == 0) continue;
int r = row + i, c = col + j;
if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
if (board[r][c] == 'M' || board[r][c] == 'X') count++;
}
}
if (count > 0) { // If it is not a 'B', stop further DFS.
board[row][col] = (char)(count + '0');
}
else { // Continue DFS to adjacent cells.
board[row][col] = 'B';
for (int i = -1; i < 2; i++) {
for (int j = -1; j < 2; j++) {
if (i == 0 && j == 0) continue;
int r = row + i, c = col + j;
if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
if (board[r][c] == 'E') updateBoard(board, new int[] {r, c});
}
}
}
}
return board;
}
}
Java: BFS
public class Solution {
public char[][] updateBoard(char[][] board, int[] click) {
int m = board.length, n = board[0].length;
Queue<int[]> queue = new LinkedList<>();
queue.add(click);
while (!queue.isEmpty()) {
int[] cell = queue.poll();
int row = cell[0], col = cell[1];
if (board[row][col] == 'M') { // Mine
board[row][col] = 'X';
}
else { // Empty
// Get number of mines first.
int count = 0;
for (int i = -1; i < 2; i++) {
for (int j = -1; j < 2; j++) {
if (i == 0 && j == 0) continue;
int r = row + i, c = col + j;
if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
if (board[r][c] == 'M' || board[r][c] == 'X') count++;
}
}
if (count > 0) { // If it is not a 'B', stop further BFS.
board[row][col] = (char)(count + '0');
}
else { // Continue BFS to adjacent cells.
board[row][col] = 'B';
for (int i = -1; i < 2; i++) {
for (int j = -1; j < 2; j++) {
if (i == 0 && j == 0) continue;
int r = row + i, c = col + j;
if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
if (board[r][c] == 'E') {
queue.add(new int[] {r, c});
board[r][c] = 'B'; // Avoid to be added again.
}
}
}
}
}
}
return board;
}
}
Python:
class Solution(object):
def updateBoard(self, board, click):
"""
:type board: List[List[str]]
:type click: List[int]
:rtype: List[List[str]]
"""
q = collections.deque([click])
while q:
row, col = q.popleft()
if board[row][col] == 'M':
board[row][col] = 'X'
else:
count = 0
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if i == 0 and j == 0:
continue
r, c = row + i, col + j
if not (0 <= r < len(board)) or not (0 <= c < len(board[r])):
continue
if board[r][c] == 'M' or board[r][c] == 'X':
count += 1 if count:
board[row][col] = chr(count + ord('0'))
else:
board[row][col] = 'B'
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if i == 0 and j == 0:
continue
r, c = row + i, col + j
if not (0 <= r < len(board)) or not (0 <= c < len(board[r])):
continue
if board[r][c] == 'E':
q.append((r, c))
board[r][c] = ' ' return board
Python:
# Time: O(m * n)
# Space: O(m * n)
class Solution2(object):
def updateBoard(self, board, click):
"""
:type board: List[List[str]]
:type click: List[int]
:rtype: List[List[str]]
"""
row, col = click[0], click[1]
if board[row][col] == 'M':
board[row][col] = 'X'
else:
count = 0
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if i == 0 and j == 0:
continue
r, c = row + i, col + j
if not (0 <= r < len(board)) or not (0 <= c < len(board[r])):
continue
if board[r][c] == 'M' or board[r][c] == 'X':
count += 1 if count:
board[row][col] = chr(count + ord('0'))
else:
board[row][col] = 'B'
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if i == 0 and j == 0:
continue
r, c = row + i, col + j
if not (0 <= r < len(board)) or not (0 <= c < len(board[r])):
continue
if board[r][c] == 'E':
self.updateBoard(board, (r, c)) return board
C++:
// Time: O(m * n)
// Space: O(m + n) class Solution {
public:
vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
queue<vector<int>> q;
q.emplace(click);
while (!q.empty()) {
int row = q.front()[0], col = q.front()[1];
q.pop();
if (board[row][col] == 'M') {
board[row][col] = 'X';
} else {
int count = 0;
for (int i = -1; i < 2; ++i) {
for (int j = -1; j < 2; ++j) {
if (i == 0 && j == 0) {
continue;
}
int r = row + i, c = col + j;
if (r < 0 || r >= board.size() || c < 0 || c < 0 || c >= board[r].size()) {
continue;
}
if (board[r][c] == 'M' || board[r][c] == 'X') {
++count;
}
}
} if (count > 0) {
board[row][col] = count + '0';
} else {
board[row][col] = 'B';
for (int i = -1; i < 2; ++i) {
for (int j = -1; j < 2; ++j) {
if (i == 0 && j == 0) {
continue;
}
int r = row + i, c = col + j;
if (r < 0 || r >= board.size() || c < 0 || c < 0 || c >= board[r].size()) {
continue;
}
if (board[r][c] == 'E') {
vector<int> next_click = {r, c};
q.emplace(next_click);
board[r][c] = 'B';
}
}
}
}
}
} return board;
}
};
C++:
// Time: O(m * n)
// Space: O(m * n)
class Solution2 {
public:
vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
int row = click[0], col = click[1];
if (board[row][col] == 'M') {
board[row][col] = 'X';
} else {
int count = 0;
for (int i = -1; i < 2; ++i) {
for (int j = -1; j < 2; ++j) {
if (i == 0 && j == 0) {
continue;
}
int r = row + i, c = col + j;
if (r < 0 || r >= board.size() || c < 0 || c < 0 || c >= board[r].size()) {
continue;
}
if (board[r][c] == 'M' || board[r][c] == 'X') {
++count;
}
}
} if (count > 0) {
board[row][col] = count + '0';
} else {
board[row][col] = 'B';
for (int i = -1; i < 2; ++i) {
for (int j = -1; j < 2; ++j) {
if (i == 0 && j == 0) {
continue;
}
int r = row + i, c = col + j;
if (r < 0 || r >= board.size() || c < 0 || c < 0 || c >= board[r].size()) {
continue;
}
if (board[r][c] == 'E') {
vector<int> next_click = {r, c};
updateBoard(board, next_click);
}
}
}
}
} return board;
}
};
All LeetCode Questions List 题目汇总
[LeetCode] 529. Minesweeper 扫雷的更多相关文章
- LN : leetcode 529 Minesweeper
lc 529 Minesweeper 529 Minesweeper Let's play the minesweeper game! You are given a 2D char matrix r ...
- LeetCode 529. Minesweeper
原题链接在这里:https://leetcode.com/problems/minesweeper/description/ 题目: Let's play the minesweeper game ( ...
- 529 Minesweeper 扫雷游戏
详见:https://leetcode.com/problems/minesweeper/description/ C++: class Solution { public: vector<ve ...
- 529. Minesweeper扫雷游戏
［抄题］: Let's play the minesweeper game (Wikipedia, online game)! You are given a 2D char matrix repre ...
- [LeetCode] Minesweeper 扫雷游戏
Let's play the minesweeper game (Wikipedia, online game)! You are given a 2D char matrix representin ...
- Week 5 - 529.Minesweeper
529.Minesweeper Let's play the minesweeper game (Wikipedia, online game)! You are given a 2D char ma ...
- leetcode笔记（七）529. Minesweeper
题目描述 Let's play the minesweeper game (Wikipedia, online game)! You are given a 2D char matrix repres ...
- Java实现 LeetCode 529 扫雷游戏(DFS)
529. 扫雷游戏 让我们一起来玩扫雷游戏! 给定一个代表游戏板的二维字符矩阵. 'M' 代表一个未挖出的地雷,'E' 代表一个未挖出的空方块,'B' 代表没有相邻(上,下,左,右,和所有4个对角线) ...
- 【LeetCode】529. Minesweeper 解题报告（Python & C++）
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 日期 题目地址:https://leetco ...
随机推荐
- 2019年牛客多校第一场 C题Euclidean Distance 暴力+数学
题目链接 传送门 题意 给你\(n\)个数\(a_i\),要你在满足下面条件下使得\(\sum\limits_{i=1}^{n}(a_i-p_i)^2\)最小(题目给的\(m\)只是为了将\(a_i\ ...
- Alpha冲刺随笔九：第九天
课程名称:软件工程1916|W(福州大学) 作业要求:项目Alpha冲刺(十天冲刺) 团队名称:葫芦娃队 作业目标:在十天冲刺里对每天的任务进行总结. 随笔汇总:https://www.cnblogs ...
- crc计算工具推荐
比较好的crc计算工具,32位64位系统都可以用的.crc的校验方法也很多.推荐使用 下载地址
- STM32F103 串口-IAP程序升级
STM32F103 串口-IAP程序升级 通常情况下我们给STM32系列的单片机烧录程序文件的时候,使用SWD.J-link或者通过设置BOOT引脚后,使用串口进行程序下载,这样的方式直接一次性将程序 ...
- 【NOIP 2017】宝藏 D2 T2
参考From 传送门 写的很清晰了 AC code: #include <bits/stdc++.h> using namespace std; const int MAXN = 12; ...
- Goexit
package main import ( "fmt" "runtime" ) func test() { defer fmt.Println("cc ...
- isntall
#! /bin/bash # zabbix server address zabbix='10.10.238.110' # repo address REPO='10.10.238.114:4507' ...
- Break关键字和Continue关键字
1.Break关键字 在循环体内,只要代码遇到break,程序立马结束当前循环. 当前循环指的是break语句所在的循环体.(直接跳到大括号外) // 例1: 1到10,第一是数字是2的倍数,控制台输 ...
- JavaScript对象及面向对象
1.创建对象(1)自定义对象 语法:var 对象名称=new Object();(2)内置对象 String(字符串)对象. Date(对象)对象 Array(数组)对象 Boll ...
- MongoDB 关系型数据库表（集合）与表（集合）之间的几种关系
简述关系数据库中表与表的 3 种关系 一对一的关系:例如:一个人对应一个唯一的身份证号,即为一对一的关系. 一对多关系 :例如:一个班级对应多名学生,一个学生只能属于一个班级,即为一对多关系 多对多关 ...