ruby - 用每种方法在Ruby中排序

我有以下代码：

def shuffle arr
  shuffled = []
  while arr.length > 0
    randNum = rand(arr.length)
    idx = 0
    unshuf = []
    arr.each do |item|
      if idx == randNum
        shuffled.push item
      else
        unshuf.push item
      end
      idx = idx + 1
    end
    arr = unshuf
  end
  return shuffled
end

puts(shuffle([ 11, 12, 13]))

我正试图理解数组unshuf的状态，因为调用方法shuffle每次调用方法unshuf之前，都会将数组each重置为空数组。在这种实现方式下，如何为下一轮的unshuf条件保留推入if的数字？

最佳答案

def shuffle arr
  shuffled = []

  while arr.length > 0
    randNum = rand(arr.length)

    idx = 0
    unshuf = []

    print '--- arr='; p arr # <--------
    arr.each do |item|
      if idx == randNum
        shuffled.push item
      else
        unshuf.push item
      end
      print 'shuffled='; print shuffled; print ', unshuf='; p unshuf # <--------
      idx = idx + 1
    end

    arr = unshuf
  end

  return shuffled
end

p shuffle([ 11, 12, 13, 14, 15])

执行：

$ ruby -w t.rb
--- arr=[11, 12, 13, 14, 15]
shuffled=[], unshuf=[11]
shuffled=[], unshuf=[11, 12]
shuffled=[13], unshuf=[11, 12]
shuffled=[13], unshuf=[11, 12, 14]
shuffled=[13], unshuf=[11, 12, 14, 15]
--- arr=[11, 12, 14, 15]
shuffled=[13, 11], unshuf=[]
shuffled=[13, 11], unshuf=[12]
shuffled=[13, 11], unshuf=[12, 14]
shuffled=[13, 11], unshuf=[12, 14, 15]
--- arr=[12, 14, 15]
shuffled=[13, 11, 12], unshuf=[]
shuffled=[13, 11, 12], unshuf=[14]
shuffled=[13, 11, 12], unshuf=[14, 15]
--- arr=[14, 15]
shuffled=[13, 11, 12, 14], unshuf=[]
shuffled=[13, 11, 12, 14], unshuf=[15]
--- arr=[15]
shuffled=[13, 11, 12, 14, 15], unshuf=[]
[13, 11, 12, 14, 15]

说明：arr的内容在无序和非无序之间随机分布。然后unshuf替换arr，如果unshuf中有什么东西，while循环会在shuffle中随机重新分配一点，在unshuf中随机重新分配一点等等，直到unsuf为空因此，shuffled是由最初给方法的参数arr中随机抽取的元素逐步填充的，但是shuffle也是通过只保留arr中的unsurf元素而逐步清空的当然，在移动到arr之后，unsuf必须重置为空数组才能接收新的分发，而shuffled必须保持增长希望一切都清楚。

关于ruby - 用每种方法在Ruby中排序，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/14095273/