blocks|key|1590401|text|+for+(Iterator<Tourist>+itr+=+tourists.iterator();+itr.hasNext();)+{
++++++if+(itr.next()+==+null)+{+itr.remove();+}
+}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1590402|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> for (Iterator&lt;Tourist&gt; itr = tourists.iterator(); itr.hasNext();) {
 if (itr.next() == null) { itr.remove(); }
 }
</code></pre>

blocks|key|1805080|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805081|tourists.removeAll(Collections.singleton(null));|code-block|syntax|javascript|1805082|阅读Java+API。代码将为不可变列表(如使用Arrays.asList创建的列表)抛出java.lang.UnsupportedOperationException；有关更多详细信息，请参阅this+answer。|offset|length|style|CODE|1805083|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/util/Collections.html#singleton-T-|1|https://stackoverflow.com/a/15381271/59087^0|0|0|O|D|19|13|2|8|0|2P|B|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]|$I|16|J|17|1|18]]|A|$]]|$1|M|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

Try:

<pre><code>tourists.removeAll(Collections.singleton(null));
</code></pre>

Read the <a href="http://docs.oracle.com/javase/8/docs/api/java/util/Collections.html#singleton-T-" rel="noreferrer">Java API</a>. The code will throw <code>java.lang.UnsupportedOperationException</code> for immutable lists (such as created with <code>Arrays.asList</code>); see <a href="https://stackoverflow.com/a/15381271/59087">this answer</a> for more details.

blocks|key|1805085|text|效率不高，但很短|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805086|while(tourists.remove(null));|code-block|syntax|javascript|1805087|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Not efficient, but short

<pre><code>while(tourists.remove(null));
</code></pre>

blocks|key|1590667|text|我尝试了一下，发现trimToSize()似乎起作用了。我在Android平台上工作，所以可能会有所不同。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1590668|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I played around with this and found out that trimToSize() seems to work. I am working on the Android platform so it might be different.

blocks|key|1590682|text|有一种简单的方法可以从collection中删除所有NULL值，.You必须将包含null的集合作为参数传递给removeAll()方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1590683|List+s1=new+ArrayList();
s1.add(null);

yourCollection.removeAll(s1);|code-block|syntax|javascript|1590684|entityMap^0|B|A|15|4|1J|B|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

There is an easy way of removing all the <code>null</code> values from <code>collection</code>.You have to pass a collection containing null as a parameter to <code>removeAll()</code> method

<pre><code>List s1=new ArrayList();
s1.add(null);

yourCollection.removeAll(s1);
</code></pre>

blocks|key|1590832|text|如果你喜欢不可变的数据对象，或者你只是不想破坏输入列表，你可以使用Guava的谓词。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1590833|ImmutableList.copyOf(Iterables.filter(tourists,+Predicates.notNull()))|code-block|syntax|javascript|1590834|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you prefer immutable data objects, or if you just dont want to be destructive to the input list, you can use Guava's predicates.

<pre><code>ImmutableList.copyOf(Iterables.filter(tourists, Predicates.notNull()))
</code></pre>

blocks|key|1805448|text|这是从arraylist中删除默认空值的简单方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805449|+++++tourists.removeAll(Arrays.asList(null));++|code-block|syntax|javascript|1805450|否则，将字符串值"null“从arraylist中移除|1805451|+++++++tourists.removeAll(Arrays.asList("null"));++|1805452|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|$]]

This is easy way to remove default null values from arraylist

<pre><code> tourists.removeAll(Arrays.asList(null)); 
</code></pre>

otherwise String value "null" remove from arraylist

<pre><code> tourists.removeAll(Arrays.asList("null")); 
</code></pre>

blocks|key|1805479|text|我们可以使用迭代器来删除所有的空值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805480|Iterator<Tourist>+itr=+tourists.iterator();
while(itr.hasNext()){
++++if(itr.next()+==+null){
++++++++itr.remove();
++++}
}|code-block|syntax|javascript|1805481|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

We can use iterator for the same to remove all the null values.

<pre><code>Iterator&lt;Tourist&gt; itr= tourists.iterator();
while(itr.hasNext()){
 if(itr.next() == null){
 itr.remove();
 }
}
</code></pre>

blocks|key|1805552|text|在2015年，这是最好的方式(Java+8)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805553|tourists.removeIf(Objects::isNull);|code-block|syntax|javascript|1805554|1805555|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

As of 2015, this is the best way (Java 8):

<pre><code>tourists.removeIf(Objects::isNull);
</code></pre>

Note: This code will throw <code>java.lang.UnsupportedOperationException</code> for fixed-size lists (such as created with Arrays.asList), including immutable lists.

blocks|key|1591071|text|Objects类有一个可与filter一起使用的nonNull+Predicate。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1591072|例如：|1591073|tourists.stream().filter(Objects::nonNull).collect(Collectors.toList());|code-block|syntax|javascript|1591074|entityMap^0|0|7|D|6|O|7|W|9|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

The <code>Objects</code> class has a <code>nonNull</code> <code>Predicate</code> that can be used with <code>filter</code>.

For example:

<pre><code>tourists.stream().filter(Objects::nonNull).collect(Collectors.toList());
</code></pre>

blocks|key|1805713|text|在Java+8之前，您应该使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805714|tourists.removeAll(Collections.singleton(null));|code-block|syntax|javascript|1805715|Java+8之后的使用：|1805716|tourists.removeIf(Objects::isNull);|1805717|1805718|这里的原因是时间复杂性。数组的问题是删除操作可能需要O(n)时间才能完成。实际上，在Java中，这是要移动的剩余元素的数组副本，以替换空白点。这里提供的许多其他解决方案都会引发此问题。前者在技术上是O(n*m)，其中m是1，因为它是一个单例null:所以O(n)|1805719|你应该对单例进行removeAll，在内部它做一个batchRemove()，它有一个读位置和一个写位置。并迭代该列表。当它遇到null时，它只是简单地将读取位置迭代1。当它们相同时，它传递，当它们不同时，它继续复制这些值。然后在最后修剪到合适的大小。|1805720|它在内部有效地做到了这一点：|1805721|public+static+<E>+void+removeNulls(ArrayList<E>+list)+{
++++int+size+=+list.size();
++++int+read+=+0;
++++int+write+=+0;
++++for+(;+read+<+size;+read%2B%2B)+{
++++++++E+element+=+list.get(read);
++++++++if+(element+==+null)+continue;
++++++++if+(read+!=+write)+list.set(write,+element);
++++++++write%2B%2B;
++++}
++++if+(write+!=+size)+{
++++++++list.subList(write,+size).clear();
++++}
}|1805722|你可以清楚地看到它是一个O(n)操作。|1805723|唯一可能更快的是，如果你从两端迭代列表，当你发现一个空值时，你设置它的值等于你在末尾找到的值，并递减那个值。并迭代，直到两个值匹配。您可能会弄乱顺序，但会大大减少您设置的值的数量，而不是您单独设置的值。这是一个很好的方法，但在这里不会有太大帮助，因为.set()基本上是免费的，但这种形式的删除对您的腰带来说是一个有用的工具。|1805724|1805725|for+(Iterator<Tourist>+itr+=+tourists.iterator();+itr.hasNext();)+{
++++++if+(itr.next()+==+null)+{+itr.remove();+}
+}|1805726|虽然这看起来很合理，但迭代器上的.remove()在内部调用：|1805727|ArrayList.this.remove(lastRet);|1805728|这也是remove中的O(n)操作。它做了一个System.arraycopy()，这也不是你想要的，如果你关心速度的话。这使其成为n%5E2。|1805729|还有：|1805730|while(tourists.remove(null));|1805731|它是O(m*n%5E2)。在这里，我们不仅迭代列表。每次匹配空值时，我们都会重复整个列表。然后，我们执行n/2+(平均)操作来执行System.arraycopy()来执行删除。从字面上讲，你可以在值项和空值项之间对整个集合进行排序，并在更短的时间内修剪结尾。事实上，所有坏掉的都是这样的。至少在理论上，实际的system.arraycopy实际上并不是一个N操作。在理论上，理论和实践是一回事，但在实践中却不是。|1805732|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1G|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1J|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1K|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1L|8|@]|9|@]|A|$]]|$1|R|3|S|5|D|7|1M|8|@]|9|@]|A|$E|F]]|$1|T|3|U|5|6|7|1N|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1O|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1R|8|@]|9|@]|A|$]]|$1|12|3|13|5|D|7|1S|8|@]|9|@]|A|$E|F]]|$1|14|3|15|5|6|7|1T|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|1U|8|@]|9|@]|A|$]]|$1|18|3|19|5|D|7|1V|8|@]|9|@]|A|$E|F]]|$1|1A|3|1B|5|6|7|1W|8|@]|9|@]|A|$]]|$1|1C|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]]|1D|$]]

Pre-Java 8 you should use:

<pre><code>tourists.removeAll(Collections.singleton(null));
</code></pre>

Post-Java 8 use:

<pre><code>tourists.removeIf(Objects::isNull);
</code></pre>

<hr>

The reason here is time complexity. The problem with arrays is that a remove operation can take O(n) time to complete. Really in Java this is an array copy of the remaining elements being moved to replace the empty spot. Many other solutions offered here will trigger this issue. The former is technically O(n*m) where m is 1 because it's a singleton null: so O(n)

You should removeAll of the singleton, internally it does a batchRemove() which has a read position and a write position. And iterates the list. When it hits a null, it simply iterates the read position by 1. When they are the same it passes, when they are different it keeps moving along copying the values. Then at the end it trims to size.

It effectively does this internally:

<pre><code>public static &lt;E&gt; void removeNulls(ArrayList&lt;E&gt; list) {
 int size = list.size();
 int read = 0;
 int write = 0;
 for (; read &lt; size; read++) {
 E element = list.get(read);
 if (element == null) continue;
 if (read != write) list.set(write, element);
 write++;
 }
 if (write != size) {
 list.subList(write, size).clear();
 }
}
</code></pre>

Which you can explicitly see is an O(n) operation.

The only thing that could ever be faster is if you iterated the list from both ends, and when you found a null, you set its value equal to the value you found at the end, and decremented that value. And iterated until the two values matched. You'd mess up the order, but would vastly reduce the number of values 
you set vs. ones you left alone. Which is a good method to know but won't help much here as .set() is basically free, but that form of delete is a useful tool for your belt.

<hr>

<pre><code>for (Iterator&lt;Tourist&gt; itr = tourists.iterator(); itr.hasNext();) {
 if (itr.next() == null) { itr.remove(); }
 }
</code></pre>

While this seems reasonable enough, the .remove() on the iterator internally calls:

<pre><code>ArrayList.this.remove(lastRet);
</code></pre>

Which is again the O(n) operation within the remove. It does an System.arraycopy() which is again not what you want, if you care about speed. This makes it n^2.

There's also:

<pre><code>while(tourists.remove(null));
</code></pre>

Which is O(m*n^2). Here we not only iterate the list. We reiterate the entire list, each time we match the null. Then we do n/2 (average) operations to do the System.arraycopy() to perform the remove.
You could quite literally, sort the entire collection between items with values and items with null values and trim the ending in less time. In fact, that's true for all the broken ones. At least in theory, the actual system.arraycopy isn't actually an N operation in practice. In theory, theory and practice are the same thing; in practice they aren't.

blocks|key|1805748|text|与@Lithium+answer类似，但不会抛出"List可能不包含类型null“错误：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805749|+++list.removeAll(Collections.<T>singleton(null));|code-block|syntax|javascript|1805750|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Similar to @Lithium answer but does not throw a "List may not contain type null" error:

<pre><code> list.removeAll(Collections.&lt;T&gt;singleton(null));
</code></pre>

blocks|key|1805752|text|List<String>+colors+=+new+ArrayList<>(
Arrays.asList("RED",+null,+"BLUE",+null,+"GREEN"));
//+using+removeIf()+%2B+Objects.isNull()
colors.removeIf(Objects::isNull);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1805753|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;String&gt; colors = new ArrayList&lt;&gt;(
Arrays.asList("RED", null, "BLUE", null, "GREEN"));
// using removeIf() + Objects.isNull()
colors.removeIf(Objects::isNull);
</code></pre>

blocks|key|1805836|text|我主要使用的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1805837|list.removeAll(Collections.singleton(null));|code-block|syntax|javascript|1805838|但在我学习了Java+8之后，我转向了这个：|1805839|List.removeIf(Objects::isNull);|1805840|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|$]]

Mainly I'm using this: 

<pre><code>list.removeAll(Collections.singleton(null));
</code></pre>

But after I learned Java 8, I switched to this:

<pre><code>List.removeIf(Objects::isNull);
</code></pre>

I try with a loop like that

<pre><code>// ArrayList tourists

for (Tourist t : tourists) {
 if (t != null) { 
 t.setId(idForm); 
 } 
}
</code></pre>

But it isn't nice. Can anyone suggest me a better solution? 

<hr>

Some useful benchmarks to make better decision: 

<a href="http://www.mkyong.com/java/while-loop-for-loop-and-iterator-performance-test-java/" rel="noreferrer">While loop, For loop and Iterator Performance Test</a>

How to remove all null elements from a ArrayList or String Array?

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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我试着用这样的循环// ArrayList touristsfor (Tourist t : tourists) {    if (t != null) {             t.setId(idForm);     }   }但这并不好。有人能给我一个更好的解决方案吗？一些有用的基准来做出更好的决定：

问如何从ArrayList或字符串数组中移除所有空元素？
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问如何从ArrayList或字符串数组中移除所有空元素？EN