如何从递归函数返回一个数组

Question

我有一个板(多维数组)。你从左上角开始，可以向右或向下移动。基本情况是右下角。

我做了一个函数来查找所有可能的移动和它们在棋盘中的值。当达到基本情况时，移动和值被存储为数组ARR_SUMNUM。

ARR_SUMNUM = []
def arr_all_moves(ary,a=0,b=0,sum_num=0,directions="")
  rows = ary.length
  cols = ary[0].length
  goal = rows-1 + cols-1
  curr_num = ary[a][b]
  sum_num += curr_num
  if [a,b] == [rows-1,cols-1]
    ARR_SUMNUM.push([sum_num, directions])
    return
  end
  if a == rows -1 
    return arr_all_moves(ary,a,b+1,sum_num,directions+="right ")
  elsif b == cols -1
    return arr_all_moves(ary,a+1,b,sum_num,directions+="down ")
  end
  arr_all_moves(ary,a,b+1,sum_num,directions+="right ")
  directions.chomp!("right ") ##### I realize this was the source of all my issues
  arr_all_moves(ary,a+1,b,sum_num,directions+="down ")
  return ARR_SUMNUM
end
board = [   [1,3,4],
            [5,6,999],
            [8,9,10],
            [11,12,13],
            [2000,42,13]
        ]
p arr_all_moves(board)

我如何让这个函数没有像ARR_SUMNUM这样的常量而使用局部变量来工作?我试图使用arr_sumnum代替，但我得到一个空数组。

Answer 1

最简单的方法可能是通过将数组设置为方法参数之一，在每次调用时传递该数组。

def arr_all_moves(ary, a=0, b=0, sum_num=0, directions=" ", arr_sum_num=[])
  rows = ary.length
  cols = ary[0].length
  goal = rows-1 + cols-1
  curr_num = ary[a][b]
  sum_num += curr_num
  if [a,b] == [rows-1,cols-1]
    arr_sum_num.push([sum_num, directions])
    return
  end
  if a == rows -1
    return arr_all_moves(ary,a,b+1,sum_num,directions+="right ", arr_sum_num)
  elsif b == cols -1
    return arr_all_moves(ary,a+1,b,sum_num,directions+="left ", arr_sum_num)
  end
  arr_all_moves(ary,a,b+1,sum_num,directions+="right ", arr_sum_num)
  arr_all_moves(ary,a+1,b,sum_num,directions+="left ", arr_sum_num)
  return arr_sum_num
end

Answer 2

有两种方法，其中一种方法已经提到过，它涉及到将数组向下传播到递归函数调用。另一种方法是向上传播数组并将它们添加到调用函数中。

def arr_all_moves(ary,a=0,b=0,sum_num=0,directions=" ")
  rows = ary.length
  cols = ary[0].length
  goal = rows-1 + cols-1
  curr_num = ary[a][b]
  sum_num += curr_num
  if [a,b] == [rows-1,cols-1]
    return [[sum_num, directions]]
  end
  if a == rows -1 
    return arr_all_moves(ary,a,b+1,sum_num,directions+="right ")
  elsif b == cols -1
    return arr_all_moves(ary,a+1,b,sum_num,directions+="left ")
  end
  return arr_all_moves(ary,a,b+1,sum_num,directions+="right ") + 
         arr_all_moves(ary,a+1,b,sum_num,directions+="left ")
end
board = [   [1,3,4],
            [5,6,999],
            [8,9,10],
            [11,12,13],
            [2000,42,13]
        ]
p arr_all_moves(board)

Answer 3

这是一个可能的解决方案，您应该使用down而不是left，因为我们正在向右或向下穿越二维数组以到达右下角。

def arr_all_moves(ary, a=0, b=0, sum_num=0, directions=" ")
  rows = ary.length
  cols = ary[0].length
  sum_num += ary[a][b]
  if [a,b] == [rows-1,cols-1]
    [] << [sum_num, directions]
  elsif a == rows-1
    arr_all_moves(ary, a, b+1, sum_num, directions+="right ")
  elsif b == cols-1
    arr_all_moves(ary, a+1, b, sum_num, directions+="down ")
  else
    arr_all_moves(ary, a, b+1, sum_num, directions+="right ") + 
    arr_all_moves(ary, a+1, b, sum_num, directions+="down ")    
  end
end
board = [  
          [1,3,4],
          [5,6,999],
          [8,9,10],
          [11,12,13],
          [2000,42,13]
        ]
p arr_all_moves(board)