blocks|key|1994430|text|List<Location>+spawnList+=+arena.getManager().getRandomSpawns();
Location+random;
Random+r+=+new+Random();

do+{
++random+=+spawnList.get(r.nextInt(spawnList.size()))
}+while(random.isOccupied());|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1994431|如果所有的位置都被占用，这将失败，你应该在之前检查这一点。|unstyled|offset|length|style|BOLD|1994432|entityMap^0|0|2|3|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>List&lt;Location&gt; spawnList = arena.getManager().getRandomSpawns();
Location random;
Random r = new Random();

do {
 random = spawnList.get(r.nextInt(spawnList.size()))
} while(random.isOccupied());
</code></pre>

This will fail if all locations are occupied, you should check this before.

blocks|key|1994451|text|您可以选择以下两种方式之一：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1994452|1994453|推送-当某个位置变为可用时，通知该位置现在可用。(通过调用一个方法，对于example).|ordered-list-item|1994454|Polling：，就像您现在正在做的那样。可以保存可用位置的集合，当某个位置变为可用时，会将其添加到该集合中。您可以等待列表具有值。我建议使用A+blockig+queue|offset|length|1994455|1994456|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/BlockingQueue.html^0|0|0|0|1Z|F|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|T|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|V|8|@]|9|@$H|W|I|X|1|Y]]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You can choose one of two ways:

<ol>
<li>Push - when a location becomes available , notify that it is now available. (By calling a method, for example).</li>
<li>Polling: Something like you are doing now. It is possible to hold a collection of available locations, when a location becomes available it is added to the collection. you can wait for the list to have values. I would suggest <a href="http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/BlockingQueue.html" rel="nofollow">A blockig queue</a>: </li>
</ol>

blocks|key|1599073|text|您可以声明一个标志来检查是否找到候选位置，并使用while+-+loop生成随机位置，例如，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1599074|++++Location+random+=+null;

boolean+foundLocation+=+false;
while(!foundLocation)
{
++++random+=+spawnList.get(new+Random().nextInt(spawnList.size()));
++++if(!random.isOccupied())
++++{
++++++++foundLocation+=+true;
++++}
}|code-block|syntax|javascript|1599075|注意:这里的假设位置列表中至少有一个位置没有被占用。如果所有的位置都被占用了。那么上面的代码就无法使用了。它将在无限循环中。我们最好先检查一下列表中是否至少有一个位置没有被占用。|offset|length|style|BOLD|1599076|entityMap^0|0|0|0|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can declare a flag to check if candidate Location is found, and using while - loop to generate random Location, like,

<pre><code> Location random = null;

boolean foundLocation = false;
while(!foundLocation)
{
 random = spawnList.get(new Random().nextInt(spawnList.size()));
 if(!random.isOccupied())
 {
 foundLocation = true;
 }
}
</code></pre>

Note: Here has a assumption that there is at least one Location in the Location List, which is not occupied. If all of the Locations are occupied. Then the above code can not be used. It will be in infinite loop. We'd better check if at least one Location is not occupied in the List first.

blocks|key|5779639|text|简单的方法是在循环中随机化一个位置，直到找到一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5779640|List<Location>+spawnList+=+arena.getManager().getRandomSpawns();+//+Returns+a+list+of+possible+locations
Location+random+=+spawnList.get(new+Random().nextInt(spawnList.size()));+//+Selects+a+random+location+from+the+list

while+(random.isOccupied())+{
++++random+=+spawnList.get(new+Random().nextInt(spawnList.size()));
}|code-block|syntax|javascript|5779641|这里的问题是，如果大多数位置已经被占用，这可能需要很长时间。|5779642|一种“更安全”的方法，无论预占位置的百分比如何，都承诺相同的性能顺序，可以将位置列表打乱，然后简单地迭代它：|5779643|List<Location>+spawnList+=+new+LinkedList<Location>(arena.getManager().getRandomSpawns());
Location+random+=+null;

for+(Location+loc+:+spawnList)+{
++++if+(!loc.isOccupied())+{
++++++++random+=+loc;
++++}
}|5779644|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The trivial approach would be to randomize a location in a loop until one is found:

<pre><code>List&lt;Location&gt; spawnList = arena.getManager().getRandomSpawns(); // Returns a list of possible locations
Location random = spawnList.get(new Random().nextInt(spawnList.size())); // Selects a random location from the list

while (random.isOccupied()) {
 random = spawnList.get(new Random().nextInt(spawnList.size()));
}
</code></pre>

The problem here is that this may take a very long time if most of the locations are already occupied.

A "safer" approach, which promises the same order of performance regardless of the percentage of pre-occupied locations could be to shuffle the list of locations, and then simply iterate through it:

<pre><code>List&lt;Location&gt; spawnList = new LinkedList&lt;Location&gt;(arena.getManager().getRandomSpawns());
Location random = null;

for (Location loc : spawnList) {
 if (!loc.isOccupied()) {
 random = loc;
 }
}
</code></pre>

blocks|key|1994594|text|不是随机地探查，直到到达一个空白点，您应该|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1994595|1994596|首先收集所有可用位置，然后|ordered-list-item|1994597|随机挑选一个空闲位置。|1994598|1994599|List<Integer>+freeLocations+=+new+ArrayList<>();
for+(int+i+=+0;+i+<+spawnList.size();+i%2B%2B)
++if+(!spawnList.get(i).isOccupied)+freeLocations.add(i);

Location+random+=+
+++spawnList.get(freeLocations.get(rnd.nextInt(freeLocations.size()));|code-block|syntax|javascript|1994600|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|R|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|S|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|T|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|U|8|@]|9|@]|A|$L|M]]|$1|N|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|O|$]]

Instead of stochastically probing until you hit an empty spot, you should

<ol>
<li>first collect all available locations and then</li>
<li>pick a random free location.</li>
</ol>

 

<pre><code>List&lt;Integer&gt; freeLocations = new ArrayList&lt;&gt;();
for (int i = 0; i &lt; spawnList.size(); i++)
 if (!spawnList.get(i).isOccupied) freeLocations.add(i);

Location random = 
 spawnList.get(freeLocations.get(rnd.nextInt(freeLocations.size()));
</code></pre>

The title is probably a bit confusing, but I don't really know how to explain this. I have a list of objects, in this case, locations, and those locations can be occupied by a player. If the selected location is already occupied, how can I try to find a new location, and continue this until non-occupied location is found?

I already know that there are 20 locations, I could manually check each and every one of those locations and see if it's occupied, but is there a better way to do this?

Here is a snippet of my code.

<pre><code>List&lt;Location&gt; spawnList = arena.getManager().getRandomSpawns(); // Returns a list of possible locations
Location random = spawnList.get(new Random().nextInt(spawnList.size())); // Selects a random location from the list

if (random.isOccupied()) {
 /* Location is occupied, find another one from the list, and continue doing this until non-occupied location is found */
}
</code></pre>

Sorry if you didn't understand, I don't know a good way of explaining this.

How to try something until available object is found

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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标题可能有点混乱，但我真的不知道如何解释这一点。我有一个对象列表，在本例中是位置，这些位置可以被玩家占用。如果所选位置已被占用，我如何尝试查找新位置，并继续此操作，直到找到未占用的位置？我已经知道有20个位置，我可以手动检查每个位置，看看是否有人占用，但有没有更好的方法呢？下面是我的代码片段。List<Location...

问如何在找到可用对象之前进行尝试
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问如何在找到可用对象之前进行尝试EN