如何找到形成给定输入的可能组合

发布于 2024-12-10 19:34:59 字数 557 浏览 0 评论 0原文

我有一个像这样的列表，例如列表名称是 output 其中有：

[[[o, g], [g, o]], [[o, g, o, d]], [[o, d]], [[t, s, n, e, e, e, n, c, s]], [[t, s, n, e, e]], [[e, n, c, s]]]

我有一个像这样的输入，比如说 input 是：

ogodtsneeencs

现在显然，>输入可以由输出形成。我尝试了 output 的 subsequences() 来查找形成 input 的可能组合，但问题是它不适用于所有 代码>输入。

谁能告诉我如何找到等于input的output组合？并且可能存储在某个列表中。

提前致谢。

原文

I have a list like this, say for example the list name is output which has:

[[[o, g], [g, o]], [[o, g, o, d]], [[o, d]], [[t, s, n, e, e, e, n, c, s]], [[t, s, n, e, e]], [[e, n, c, s]]]

And I have a input like this, say input is:

ogodtsneeencs

Now obviously, the input can be formed from output. I tried the subsequences() of output to find the possible combinations that form the input, but the thing is it wont work for all the input.

Can anyone say me how I can find the combinations of output that will be equal to input? And possibly store in some list.

Thanks in advance.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

樱花细雨 2024-12-17 19:34:59

鉴于您提供的这一小组测试数据，我想出了这个：

def list = [[['o', 'g'], ['g', 'o']], [['o', 'g', 'o', 'd']], [['o', 'd']], [['t', 's', 'n', 'e', 'e', 'e', 'n', 'c', 's']], [['t', 's', 'n', 'e', 'e']], [['e', 'n', 'c', 's']]]

// For every combination of the lists
def result = list.combinations().collect { combination ->
  // Join them into strings
  combination*.join().with { stringcombo ->
    // Then find every string in the list
    stringcombo.findAll { word ->
      // Which is not a substring of another string in the list
      (stringcombo - word).every { it.indexOf( word ) == -1 }
    }
  }.permutations()*.join() // Then get every String permutation of these remaining strings
}.flatten().unique() // and get them into a single unique list

// And print them out
result.each {
  println it
}

打印出：

ogodtsneeencs
tsneeencsogod

没有更多数据，很难判断它是否正确，但这可能是您

编辑

更新以返回所有内容的一个很好的起点有效标记的排列

Given just this small set of test data you have supplied, I came up with this:

def list = [[['o', 'g'], ['g', 'o']], [['o', 'g', 'o', 'd']], [['o', 'd']], [['t', 's', 'n', 'e', 'e', 'e', 'n', 'c', 's']], [['t', 's', 'n', 'e', 'e']], [['e', 'n', 'c', 's']]]

// For every combination of the lists
def result = list.combinations().collect { combination ->
  // Join them into strings
  combination*.join().with { stringcombo ->
    // Then find every string in the list
    stringcombo.findAll { word ->
      // Which is not a substring of another string in the list
      (stringcombo - word).every { it.indexOf( word ) == -1 }
    }
  }.permutations()*.join() // Then get every String permutation of these remaining strings
}.flatten().unique() // and get them into a single unique list

// And print them out
result.each {
  println it
}

Which prints out: