動機

dfs去點格子

Problem

Let's play the minesweeper game (Wikipedia, online game)!

You are given an m x n char matrix board representing the game board where:

    t
  • 'M' represents an unrevealed mine,
    • t
  • 'E' represents an unrevealed empty square,
    • t
  • 'B' represents a revealed blank square that has no adjacent mines (i.e., above, below, left, right, and all 4 diagonals),
    • t
  • digit ('1' to '8') represents how many mines are adjacent to this revealed square, and
    • t
  • 'X' represents a revealed mine.

You are also given an integer array click where click = [clickr, clickc] represents the next click position among all the unrevealed squares ('M' or 'E').

Return the board after revealing this position according to the following rules:

    t
  1. If a mine 'M' is revealed, then the game is over. You should change it to 'X'.
    2. t
  2. If an empty square 'E' with no adjacent mines is revealed, then change it to a revealed blank 'B' and all of its adjacent unrevealed squares should be revealed recursively.
    3. t
  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. t
  4. Return the board when no more squares will be revealed.

 

Example 1:

Input: board = [["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: board = [["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"]]

 

Constraints:

    t
  • m == board.length
    • t
  • n == board[i].length
    • t
  • 1 <= m, n <= 50
    • t
  • board[i][j] is either 'M', 'E', 'B', or a digit from '1' to '8'.
    • t
  • click.length == 2
    • t
  • 0 <= clickr < m
    • t
  • 0 <= clickc < n
    • t
  • board[clickr][clickc] is either 'M' or 'E'.

Sol

在空白時用dfs去點其他格子

def surroundMine(B,i,j):
    poses = [(i-1,j-1), (i-1,j), (i-1,j+1),
             (i,j-1),           (i,j+1),
             (i+1,j-1), (i+1,j),(i+1,j+1)]
    return sum([1 if B[a][b] == 'M' else 0 for (a,b) in poses if 0 <= a and a < len(B) and 0 <= b and b < len(B[a])])
def dfs(B,i,j):
    if 0 <= i and i < len(B) and 0 <= j and j < len(B[i]):
        if B[i][j] == 'E':
            num = surroundMine(B,i,j)
            if num == 0:
                B[i][j] = 'B'
                poses = [(i-1,j-1), (i-1,j), (i-1,j+1),
                         (i,j-1),           (i,j+1),
                         (i+1,j-1), (i+1,j),(i+1,j+1)]
                [dfs(B,a,b) for (a,b) in poses]
            else:
                B[i][j] = str(num)
class Solution:
    def updateBoard(self, B: List[List[str]], c: List[int]) -> List[List[str]]:
        if B[c[0]][c[1]] != 'M':
            dfs(B,c[0],c[1])
        else:
            B[c[0]][c[1]] = 'X'
        return B