Question

[1,2,3,4,5]
=>1,2,3,4,5,4,3,2,1
=>1,2,3,2,3,4,5,4,3 #I need to be able to reverse the iteration at certain points

我首先尝试过类似的事情：

a = [1,2,3,4,5]
a.each {|i|
  if i % 9 == 0
  a.reverse!
}

但这只是反转整个数组并从它停止的索引开始计数。我需要改变each的方向，可以这么说。

Answer 1

i, counter = 0, 1                   # initialize index to 0, counter to 1 
while(i < a.length && i >= 0) do
  puts a[i]
  i+= counter                       # increment counter
  counter*= -1 if(condition)        # Multiply counter with -1 to reverse it
end

Answer 2

您可以使用Enumerator类来创建可以通过数组提供自定义迭代的自定义枚举。在下面的代码中，为了方便起见，我正在修补Array类（也因为方法与Array#cycle的相似性），尽管解决方案也可以在没有猴子修补的情况下完成。

class Array
  def reversible_cycle
    Enumerator.new do |y|
      index = 0
      direction = :forward
      loop do
        direction = :backward if index + 1 >= size
        direction = :forward if index <= 0
        y << self[index]
        index += (direction == :forward ? +1 : -1)
      end
    end
  end
end


p [1,2,3,4,5].reversible_cycle.take(9)
#=> [1, 2, 3, 4, 5, 4, 3, 2, 1]

p [1,2,3,4,5].reversible_cycle.take(13)
#=> [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5]

p [1,2,3,4,5].reversible_cycle.take(17)
#> [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1]

p [1,2,3,4,5].reversible_cycle.take(21)
#=> [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5]

对于在不向一个方向完全迭代数组的情况下改变方向的场景，您必须提供一些示例，以便可以看到如何修改上述代码以适应

Answer 3

嗯，这是你阵列的移动“光标”：

module Cursor
  def current_index
    @current_index ||= 0
  end

  def step
    @current_index = current_index + direction
    handle_boundary
  end

  def step_back
    @current_index = current_index + (direction * -1)
    handle_boundary
  end

  def handle_boundary
    if current_index == length || current_index == 0
      turn_around
    end
  end

  def direction
    @direction ||= 1
  end

  def turn_around
    @direction = direction * -1
  end

  def current
    self[current_index]
  end
end

以下是你如何使用它：

array = [1,2,3,4,5]
arary.extend Cursor
array.current     # returns the item in current position
array.step        # moves a step forward, turns around when it reaches either end of the array
array.step_back   # moves a step backward without switching the direction
array.turn_around # switch the direction

现在你可以随意旅行：D

Answer 4

你可以使用Ruby不太受欢迎的flip-flop operator。

arr = [1,2,3,4,5]

sz = arr.size
(2*sz-1).times { |i| puts i==0..i==arr.size-1 ? arr[i] : arr[sz-i-2] }
1
2
3
4
5
4
3
2
1

向前迭代然后向后迭代

4 个答案: