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minesweeper.py
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minesweeper.py
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# Time: O(m * n)
# Space: O(m + n)
# 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']]
#
# 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']]
#
# 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.
import collections
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
# 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