問題：
Yesterday I was pairing the socks from the clean laundry and figured out the way I was doing it is not very efficient.昨天我把洗乾淨的衣服的襪子配對，發現我這樣做的方式效率不高。 I was doing a naive search — picking one sock and "iterating" the pile in order to find its pair.我正在做一個天真的搜索 - 挑選一隻襪子並“迭代”一堆以找到它的一雙。 This requires iterating over n/2 * n/4 = n 2 /8 socks on average.這需要平均迭代 n/2 * n/4 = n 2 /8 個襪子。
As a computer scientist I was thinking what I could do?作爲一名計算機科學家，我在想我能做什麼？ Sorting (according to size/color/...) of course came to mind to achieve an O(NlogN) solution.排序（根據大小/顏色/...）當然會想到實現 O(NlogN) 解決方案。
Hashing or other not-in-place solutions are not an option, because I am not able to duplicate my socks (though it could be nice if I could).散列或其他非就地解決方案不是一種選擇，因爲我無法複製我的襪子（儘管如果可以的話可能會很好）。
So, the question is basically:所以，問題基本上是：
Given a pile of n pairs of socks, containing 2n elements (assume each sock has exactly one matching pair), what is the best way to pair them up efficiently with up to logarithmic extra space?給定一堆n雙襪子，包含2n元素（假設每隻襪子正好有一對匹配），用最多對數額外空間將它們有效配對的最佳方法是什麼？ (I believe I can remember that amount of info if needed.) （如果需要，我相信我能記住那麼多信息。）
I will appreciate an answer that addresses the following aspects:我將感謝解決以下方面的答案：
- A general theoretical solution for a huge number of socks.大量襪子的通用理論解決方案。
- The actual number of socks is not that large, I don't believe my spouse and I have more than 30 pairs.襪子的實際數量並沒有那麼大，我不相信我和我的配偶有30多雙。 (And it is fairly easy to distinguish between my socks and hers; can this be used as well?) （而且很容易區分我的襪子和她的襪子；這也可以使用嗎？）
- Is it equivalent to the element distinctness problem ?是否等價於元素區分問題？
解決方案：
參考一： https://stackoom.com/question/yUED參考二： How can I pair socks from a pile efficiently?