在ruby中使用类似的if条件减少重复

当实现像迷宫问题这样的递归函数时。我将尝试移动到4个方向，每一步看起来都很相似

if current_y+1<@y_length and @map[current_y+1][current_x].undected?
    path.push current_point
    next_point=  @map[current_y+1][current_x]       
    flag=DFS(next_point,target_point,path)
    if flag=='found'
        return flag
    end
end
if current_y-1>=0 and @map[current_y-1][current_x].undected?
    path.push current_point
    next_point= @map[current_y-1][current_x]
    flag=DFS(next_point,target_point,path)
    if flag=='found'
        return flag
    end
end
if current_x+1<@x_length and @map[current_y][current_x+1].undected?
    path.push current_point
    next_point=@map[current_y][current_x+1]
    flag=DFS(next_point,target_point,path)
    if flag=='found'
        return flag
    end
end
if current_x-1>=0 and @map[current_y][current_x-1].undected?
    path.push current_point
    next_point=@map[current_y][current_x-1]
    flag=DFS(next_point,target_point,path)
    if flag=='found'
        return flag
    end
end

---

如何使其尽可能短？

您可以尝试以下方法：

path.push current_point
if current_y+1<@y_length && @map[current_y+1][current_x].undected?
  next_point=  @map[current_y+1][current_x]
elsif current_y-1>=0 and @map[current_y-1][current_x].undected?
  next_point= @map[current_y-1][current_x]
elsif current_x+1<@x_length and @map[current_y][current_x+1].undected?
  next_point=@map[current_y][current_x+1]
elsif current_x-1>=0 and @map[current_y][current_x-1].undected?
  next_point=@map[current_y][current_x-1]
end
flag=DFS(next_point,target_point,path)
if flag=='found'
  return flag
end

---

而一个简单的列表加上一个each语句就行不通了

    directions = [
    [current_y+1<@y_length,current_y+1,current_x],
    [current_y-1>=0, current_y-1, current_x],
    [current_x+1<@x_length,current_y, current_x+1],
    [current_x-1>=0,current_x-1]
]

directions.each do |check_condition, y, x|

  next if @map[y][x].undected?
  next if check_condition

  path.push(current_point)
  next_point=@map[y][x]
  flag=DFS(next_point,target_point,path)

  return flag if flag=='found'

end

---

对不起，我忘了强调它是一个递归函数，这意味着每个if分支都必须通过。谢谢，现在我通过将method对象放入数组并通过它来完成它，与您的非常类似。我喜欢你的解决方案：