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.
掃雷游戲，經典的搜索問題。
1. 當點擊到雷('M')時，標記為'X'，結束搜索，游戲結束。
2. 當點擊到空方塊('E')時，如果周圍有雷，就計算雷的個數，標記為這個數字，不在搜索。如果周圍沒有雷的話， 標記為'B'，繼續搜索8個挨着的方塊。
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;
}
};
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