blocks|key|340836|text|可以从数组中获得一个HashSet，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|340837|char[]+array+=+new[]+{+'a',+'a',+'b',+'c',+'c'+};
HashSet<char>+hashSet+=+new+HashSet<char>(array);|code-block|syntax|javascript|340838|这将是一个更好的方法，比比较和发现重复的一个人。|340839|entityMap^0|A|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can get a <code>HashSet</code> from an array like:

<pre><code>char[] array = new[] { 'a', 'a', 'b', 'c', 'c' };
HashSet&lt;char&gt; hashSet = new HashSet&lt;char&gt;(array);
</code></pre>

It would be a better approach than comparing and detecting duplicates by yourself.

blocks|key|684088|text|如果你说的是简单的焦炭，我认为你可以做一些简单的事情，比如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|684089|bool[]+map+=+new+bool[256];|code-block|syntax|javascript|684090|对于元素访问：|684091|map[(int)'a'];|684092|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you are talking about simple char I think you can go with something as simple as:

<pre><code>bool[] map = new bool[256];
</code></pre>

And for element access:

<pre><code>map[(int)'a'];
</code></pre>

blocks|key|341768|text|如果您真的很担心对此进行优化，那么您可以为您的字符使用一个查找表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|341769|var+lookup+=+Enumerable.Repeat(true,+256).ToArray();
var+otherCharacters+=+HashSet<char>();|code-block|syntax|javascript|341770|然后，您可以使用对“小”字符的查找，在找到时将其翻转到true，并将otherCharacters用于unicode+.|offset|length|style|CODE|341771|就像这样：|341772|foreach+(var+c+in+myListOfChars)
{
++++try
++++{
++++++++if+(!lookup[(int)c])+{+//+do+something+}
++++++++lookup[(int)c]+=+true;
++++}
++++catch+(IndexOutOfRangeException+e)
++++{
++++++++if+(!otherCharacters.Contains(c))+{+//+do+something+}
++++++++otherCharacters.Add(c);
++++}
}|341773|对于在查找表范围之外的字符，这将遇到一个慢位，这取决于您的区域设置是否可以接受。对于基于拉丁文的字符集，这应该很好！|341774|现在..。并不是所有的世界都是在阿拉伯语/拉丁语-1范围内工作的.通过阿拉伯语文本将需要一个不同的范围。|341775|编辑：Um.我刚检查了GetHashCode()输出的数字..。好吧..。原来int的哈希码是int本身.所以优化我们的查找表可能是愚蠢的..。接下来我要检查HashSet的实现.|BOLD|341776|entityMap^0|0|0|R|4|Y|F|0|0|0|0|0|0|3|B|D|13|3|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|18|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|19|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1A|8|@$I|1B|J|1C|K|W]|$I|1D|J|1E|K|L]|$I|1F|J|1G|K|L]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|Y|$]]

If you're really worried about optimizing this, then you can use a lookup table for your characters:

<pre><code>var lookup = Enumerable.Repeat(true, 256).ToArray();
var otherCharacters = HashSet&lt;char&gt;();
</code></pre>

then, you can use the lookup for the "small" characters, flipping it to <code>true</code>when found, and use the <code>otherCharacters</code> for the unicode stuff...

Something like this:

<pre><code>foreach (var c in myListOfChars)
{
 try
 {
 if (!lookup[(int)c]) { // do something }
 lookup[(int)c] = true;
 }
 catch (IndexOutOfRangeException e)
 {
 if (!otherCharacters.Contains(c)) { // do something }
 otherCharacters.Add(c);
 }
}
</code></pre>

This will hit a slow bit for characters outside the range of the lookup table, which depending on your locale is either acceptable or not. For latin based character sets this should work fine!

Now... not all the world works in the ascii / latin-1 range... going through texts in arabic will require a different range.

EDIT: Um... I've just checked the output of <code>GetHashCode()</code> for numbers... well... it turns out that the hashcode for an <code>int</code> is the int itself... so optimizing with our lookup table is probably just stupid... I'm going to check the implementation of HashSet next...

blocks|key|341789|text|如果您只关心ASCII，那么最好的方法是一个大小为128的数组并转换为int。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|341790|+boolean[]+array+=+new+bolean[128];
+char+c+=+'a';
+array[(int)+c]+=+true;+|code-block|syntax|javascript|341791|任何更大的编码，明确地使用地图，我认为。|341792|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you are only concerned with ASCII then the best approach would be an array of size 128 and casting to an int. 

<pre><code> boolean[] array = new bolean[128];
 char c = 'a';
 array[(int) c] = true; 
</code></pre>

Any sort of larger encoding, definatly just use a map i would think.

I have an array of <code>characters</code> which I will iterate over. Once I find a character which I have not found before, I do something.

This means I need to keep track of which characters I have already met. My first choice was the <code>HashSet</code>, but I am not sure if this is the right choice, since <code>hashing</code> a single character might take longer than just <code>comparing</code> two characters. I wonder if this is true.

<ol>
<li>Is HashSet the right choice for this, or are there better options which, for example use a very small hash, or none at all.</li>
</ol>

Clarification dump

The array is actually a two dimentional array, which I receive from a function written by a college. I need to locate the position of each character too. The position of which character of a certain type is not relevant, as long as the function is not called twice for a type of character.

What I need to know is all unique characters in the multidimentional array, and for each unique character the location of one of these.

HashSet for unique characters

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个characters数组，我将对其进行迭代。一旦我找到了一个我以前没有发现过的角色，我就会做一些事情。这意味着我需要跟踪我已经见过的角色。我的第一个选择是HashSet，但我不确定这是否是正确的选择，因为hashing一个字符可能会比comparing两个字符花费更长的时间。我不知道这是不是真的。HashSet...

问唯一字符的HashSet
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