题目描述：
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.
题目大意：
扫雷游戏。
给定2维字符矩阵board，'M'表示未挖到的地雷，'E'表示未挖到的空地。'B'表示已经挖到的、四周（上下左右以及对角线）没有地雷的空地。数字'1'到'8'表示四周包含相应数字的地雷，'X'表示挖到地雷。
给定下一次鼠标点击的位置（只在'M'或者'E'处点击），根据如下规则返回点击之后的board：
如果挖到'M'，则将其变成'X'，游戏结束
如果挖到四周没有地雷的'E'，将其变成'B'，并递归地挖开其相邻无雷区域
如果挖到四周包含地雷的'E'，将其替换为四周地雷的数目
注意：
- 输入矩阵的高度和宽度范围是[1, 50]。
- 点击位置只会在'M'和'E'上，换言之，board中至少包含一个可以点击的方块。
- 输入board不会处在游戏结束的状态。
- 为了简化，未提到的规则可以忽略。例如，你不需要在游戏结束时翻开所有地雷，也不需要考虑是不是已经赢得游戏，或者在某些方块上插上了旗帜。
解题思路：
广度优先搜索（Breadth First Search）
Python代码：
class Solution(object):
def updateBoard(self, board, click):
"""
:type board: List[List[str]]
:type click: List[int]
:rtype: List[List[str]]
"""
w, h = len(board), len(board[0])
def countBoard(i, j):
cnt = 0
for di in (-1, 0, 1):
for dj in (-1, 0, 1):
ni, nj = i + di, j + dj
if ni < 0 or ni >= w or nj < 0 or nj >= h:
continue
if board[ni][nj] == 'M':
cnt += 1
return str(cnt) if cnt else 'B'
cx, cy = click
if board[cx][cy] == 'M':
board[cx][cy] = 'X'
return board
q = [click]
board[cx][cy] = countBoard(cx, cy)
if board[cx][cy] != 'B':
return board
while q:
ti, tj = q.pop(0)
for di in (-1, 0, 1):
for dj in (-1, 0, 1):
ni, nj = ti + di, tj + dj
if ni < 0 or ni >= w or nj < 0 or nj >= h:
continue
if board[ni][nj] == 'E':
board[ni][nj] = countBoard(ni, nj)
if board[ni][nj] == 'B':
q.append((ni, nj))
return board
本文链接：http://bookshadow.com/weblog/2017/02/26/leetcode-minesweeper/
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