Question

我正在做代码拼车的挑战：扫雷说明：

我的代码如下：

def minesweeper(matrix):

    for x in range(len(matrix)):
        matrix[x].insert(0, "x")
        #matrix[x].insert(len(matrix)+2, "x")
    frame = ["x" for i in range(len(matrix[0]))]
    matrix.insert(0, frame)
    matrix.insert(len(matrix), frame)

    output_matrix = [[0 for j in range(len(matrix[0]))] for i in range(len(matrix))]

    for i in range(0,len(matrix[0])-1):
        for j in range(0,len(matrix)-1):
            if matrix[i][j] == True:
                output_matrix[i][j+1] += 1 # one right
                output_matrix[i+1][j] += 1 # one down
                output_matrix[i][j-1] += 1 # one left
                output_matrix[i-1][j] += 1 # one up
                output_matrix[i+1][j+1] += 1 # one down, one right
                output_matrix[i+1][j-1] += 1 # one down, one right
                output_matrix[i-1][j+1] += 1 # one up, one right
                output_matrix[i-1][j-1] +=1 # one up, one left

    output_matrix.pop(0)
    output_matrix.pop(len(output_matrix)-1)

    for y in range(len(output_matrix)):
        output_matrix[y].pop(0)
        #output_matrix[y].pop(len(output_matrix))
    return output_matrix

＆＃34; x＆＃34;创建的边框，正如codefight用户所建议的那样，是为了确保如果我的是在矩阵的边界，炸弹数量不会转移到另一个侧。
此代码可正常工作，直到炸弹位于矩阵的最后一列，例如：

如果输入：

[[False, False, True],
 [False, False, False],
 [False, False, False]]

输出为：

[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

有人能解释清楚为什么会这样吗？
如果有人可以建议更好的方法来完成这项任务，我会很感激。
提前谢谢。

Answer 1

我刚刚颠倒了你的逻辑：我遍历输出字段并从矩阵中添加值。请注意使用例外（这是关于“x”的提示）。使用此解决方案，您不必使用pop（）缩小结果。

import itertools 

def minesweeper(matrix):

    #create the output matrix first to preserve the size
    #underscored variables to prevent warnings about unused variables
    output_matrix = [[0 for _j in range(len(matrix[0]))] for _i in range(len(matrix))]

    #unchanged
    for x in range(len(matrix)):
        matrix[x].insert(0, "x")
        matrix[x].insert(len(matrix)+2, "x")
    frame = ["x" for i in range(len(matrix[0]))]
    matrix.insert(0, frame)
    matrix.insert(len(matrix), frame)

    #to the logics the other way round: count the bombs around the output fields.
    #neighyours defines the offsets of all neighouring fields 
    neighbours = [(-1, -1), (-1, 0), (-1, 1), 
                  ( 0, -1),          ( 0, 1), 
                  ( 1, -1), ( 1, 0), ( 1, 1)] 
    for i, j in itertools.product(range(len(output_matrix[0])), range(len(output_matrix))):
        #get all indices; you could use two for-loops instead of itertools.product...
        print(i, j) # just to see how it works... please remove for final version
        for offset_i, offset_j in neighbours:
            print("   ", offset_i, offset_j ) # just to see how it works... please remove for final version
            # the exceptions do the magic here: If you add an "x", a TypeError is raised.
            # So you don't do anythithing if this happens. Otherwise you'll add 0 or 1 (adding "True" adds 1, "False" adds 0)
            try:
                output_matrix[j][i] += matrix[j + offset_j + 1][i + offset_i + 1]
                print("result = ", output_matrix[j][i]) # just to see how it works... please remove for final version
            except TypeError:
                print("pass") # just to see how it works... please remove for final version
                pass

    return output_matrix

matrix = [[False, False, True], #renamed input variable since "input" is a function name...
 [False, False, False],
 [False, False, False]]

print(minesweeper(matrix))

通常，您的解决方案正在运行（如果您取消注释行#matrix[x].insert(len(matrix)+2, "x")），但您在pop（）序列中出错。您可以使用2D切片（请参阅corresponding stackoverflow topic）并执行

output_matrix = [output_matrix[i][1:len(output_matrix)-1] for i in range(1, len(output_matrix)-1)]

代替所有pop（） - 步骤。

Codefights，扫雷，python，代码几乎正常工作

1 个答案: