Question

我正在尝试创建一个函数，该函数将从列表中的列表中获取对角线，也就是我的单词搜索难题的矩阵。

def get_left_up (start_x, start_y, board):
word = []
for row in range (start_x, -1,-1):
    for col in range (start_y,-1, -1):
        word.append(board[row][col])
word = ''.join(word)
return word

get_left_down (start_x, start_y, board):
word = []
for row in range (start_x,len(board)):
    for col in range (start_y,-1,-1):
        word.append(board[row][col])
word = ''.join(word)
return word

def get_right_up (start_x, start_y, board):
word= []
for row in range (start_x,-1,-1):
    for col in range (start_y,len(board)):
        word.append(board[row][col])
word = ''.join(word)
return word

def get_right_down (start_x, start_y, board):
word = []
for row in range (start_x, len(board)):
    for col in range (start_y,len(board)):
      word.append(board[row][col])
word = ''.join(word)
return word

到现在为止，我应该在其中工作的功能。但是，出了点问题。我打算输入与要搜索的单词匹配的第一个字母的坐标的参数，并将其从对角线的各个方向连接起来。

Answer 1

对于右下角的对角线，您需要在同一迭代中增加x-和y-，而不是遍历特定行的所有列条目。

def get_right_down (start_x, start_y, board):
   word = []
   col = start_y
   for row in range (start_x, len(board)):
      if col >= len(board):
         break
      word.append(board[row][col])
      col = col + 1
   return ''.join(word)

获取此矩阵的对角线

1 个答案: