如何对列表进行混排？ [英] How to unshuffle a list?

问题描述

我正在使用 Collections.shuffle（list）;
来整理列表，但是我不知道如何去整理它？我在考虑保存重新整理列表，然后重新整理列表，以便可以维护备份并可以在需要时恢复它，但这似乎是一种低效的方式，并且会占用时间和内存....如果您知道更合理的方法这样做，请您详细说明一下吗？
就是我的应用程序的外观：D

I am using Collections.shuffle(list); to shuffle a list but I don't know how to unshuffle it?I was thinking of saving the list before shuffling and then shuffle it so that a backup is maintained and can re restored whenever required ,but this seems like a inefficient way of doing it and will take up time and memory ....if you know a more logical way of doing it,can you please elaborate it ?? by the way here is how my app looks :D

推荐答案

没有这样的改组概念-在洗完一副纸牌后，您将如何例如，返回到先前的状态？

There's no such concept of unshuffling - after you've shuffled a deck of cards, how would you get back to the previous state, for example?

如果您以某种确定的方式订购了原始收藏，请再次对其进行排序。否则（例如，如果是手工订购的），您必须先进行复印，然后再洗。

If your original collection is ordered in some definitive way, just sort it again. Otherwise (e.g. if it's been hand-ordered) you must take a copy before you shuffle.

理论上您可以 ：

• 生成随机种子并记住它


• 创建随机，然后将其传递到 shuffle


• 稍后，创建 ArrayList<整数> 从0到大小（不包括大小）


• 使用新的 Random 随机排序列表原始种子


• 使用结果计算出每个项目的原始索引（因为您知道每个原始项目在改组后的列表中的位置）

• Generate a random seed and remember it
• Create a Random and pass that into shuffle
• Later, create an ArrayList<Integer> from 0 to size (exclusive)
• Shuffle that list with a new Random created with the original seed
• Use the results to work out the original index of each item (because you know where each original item ended up in the shuffled list)

...但这工作量很大。除非您的集合真的太大而不能保留多余的副本（别忘了它只是引用的副本，而不是整个对象的副本），否则我只会在拖曳之前克隆该集合。

... but that's an awful lot of work. Unless your collection is really too big to keep the extra copy (don't forget it's just a copy of references, not whole objects), I'd just clone the collection before shuffling.

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