在python中递归退出矩形

Question

我尝试编写一个递归函数，它接受一组整数数组并返回从左上角到右下角的路径，只在相邻单元格之间传递，其中每个新单元格必须比前一个单元格大很多.

为了获得准确的路线，我需要按以下顺序检查四个相邻的单元格：右、下、左和上。

def exit_rectangle_position(array, row, col, l):
    if col + 1 < len(array) and array[row][col+1] > array[row][col]:
        l.append([row,col+1])
        exit_rectangle_position(array, row, col+1, l)
    elif row + 1 < len(array) and array[row+1][col] > array[row][col]:
        l.append([row+1,col])
        exit_rectangle_position(array, row+1, col, l)
    elif col - 1 >= 0 and array[row][col-1] > array[row][col]:
        l.append([row,col-1])
        exit_rectangle_position(array, row, col-1, l)
    elif row - 1 >= 0 and array[row - 1][col] > array[row][col]:
        l.append([row-1,col])
        exit_rectangle_position(array, row-1, col, l)



def exit_rectangle(array):
    l = []
    l.append([0,0])
    exit_rectangle_position(array,0,0,l)
    if [len(array)-1, len(array)-1] in l:
        return l
    return []

问题是当我被卡住时，我不知道如何从我开始的地方回来。 例如数组

print(exit_rectangle([[1,2,3],[2,0,5],[3,4,5]]))

我必须回去

# [[0, 0], [1, 0], [2, 0], [2, 1], [2, 2]]

但我明白了

# []

Answer 1

递归 DFS 在这里效果很好：

def exit_rectangle(a, x=None, y=None, seen=None):
  if x is None:
    x = 0
  if y is None:
    y = 0
  if seen is None:
    seen = set()

  if x + 1 == len(a[0]) and y + 1 == len(a):
    # Exit found
    return [(x, y)]

  # Maybe we've been here before
  if (x, y) in seen:
    return None
  seen.add((x, y))

  # Go right
  if x + 1 < len(a[0]) and a[y][x] < a[y][x+1]:
    path = exit_rectangle(a, x+1, y, seen)
    if path is not None:
      return [(x, y)] + path

  # Go left
  if 0 < x and a[y][x] < a[y][x-1]:
    path = exit_rectangle(a, x-1, y, seen)
    if path is not None:
      return [(x, y)] + path

  # Go up
  if 0 < y and a[y][x] < a[y-1][x]:
    path = exit_rectangle(a, x, y-1, seen)
    if path is not None:
      return [(x, y)] + path

  # Go down
  if y + 1 < len(a) and a[y][x] < a[y+1][x]:
    path = exit_rectangle(a, x, y+1, seen)
    if path is not None:
      return [(x, y)] + path

  # Dead end
  return None


print(exit_rectangle([[1,2,3],[2,0,5],[3,4,5]]))