blocks|key|869605|text|最简单的方法是遍历。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869606|List<String>+strings+=+new+ArrayList<String>();

//+populate+strings

StringBuilder+builder+=+new+StringBuilder();

for(String+string+:+strings)+{
++++builder.append(string).append(',');
}

if(builder.length()+>+0)+{
builder.deleteCharAt(builder.length()+-+1);
}

System.out.println(builder);|code-block|syntax|javascript|869607|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|$]]

Easiest is to loop through.

<pre><code>List&lt;String&gt; strings = new ArrayList&lt;String&gt;();

// populate strings

StringBuilder builder = new StringBuilder();

for(String string : strings) {
 builder.append(string).append(',');
}

if(builder.length() &gt; 0) {
builder.deleteCharAt(builder.length() - 1);
}

System.out.println(builder);
</code></pre>

blocks|key|652370|text|您可以遍历列表并创建字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|652371|String+getStringRepresentation(ArrayList<Character>+list)
{++++
++++StringBuilder+builder+=+new+StringBuilder(list.size());
++++for(Character+ch:+list)
++++{
++++++++builder.append(ch);
++++}
++++return+builder.toString();
}|code-block|syntax|javascript|652372|将StringBuilder的容量设置为列表大小是一个重要的优化。如果不这样做，一些append调用可能会触发构建器的内部调整大小。|offset|length|style|CODE|652373|另外，toString()返回一个人类可读的ArrayList内容格式。不值得花时间从其中过滤掉不必要的字符。它的实现明天可能会改变，你将不得不重写你的过滤代码。|652374|entityMap^0|0|0|1|D|16|6|0|3|A|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

You can iterate through the list and create the string.
<pre><code>String getStringRepresentation(ArrayList&lt;Character&gt; list)
{ 
 StringBuilder builder = new StringBuilder(list.size());
 for(Character ch: list)
 {
 builder.append(ch);
 }
 return builder.toString();
}
</code></pre>
Setting the capacity of the <code>StringBuilder</code> to the list size is an important optimization. If you don't do this, some of the <code>append</code> calls may trigger an internal resize of the builder.
As an aside, <code>toString()</code> returns a human-readable format of the ArrayList's contents. It is not worth the time to filter out the unnecessary characters from it. It's implementation could change tomorrow, and you will have to rewrite your filtering code.

blocks|key|869696|text|有许多可用的解决方案。您可以遍历字符并将其附加到StringBuilder，然后在完成附加后，在StringBuilder上调用.toString。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869697|或者使用apache+commons-lang项目中的commons-lang+StringUtils.join。|869698|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Many solutions available. You can iterate over the chars and append to a StringBuilder, then when finished appending, call .toString on the StringBuilder.

Or use something like commons-lang StringUtils.join from the apache commons-lang project.

blocks|key|652441|text|我会说：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|652442|public+String+arayListToString(ArrayList+arrayList){

StringBuffer+b+=+new+StringBuffer();

for(String+s+:+arrayList){
+++b.append(s);
+++b.append(",");
}

return+b.toString();
}|code-block|syntax|javascript|652443|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|$]]

I would say : 

<pre><code>public String arayListToString(ArrayList arrayList){

StringBuffer b = new StringBuffer();

for(String s : arrayList){
 b.append(s);
 b.append(",");
}

return b.toString();
}
</code></pre>

blocks|key|869758|text|您可以使用toString()和RegExp来完成，而不需要任何循环和流：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869759|List<Character>+list+=+Arrays.asList('a',+'b',+'c');+String+s+=+list.toString().replaceAll("[,\\s\\[\\]]",+"");|offset|length|style|CODE|869760|entityMap^0|0|0|33|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

You can do it using toString() and RegExp without any loops and streams:

<code>List&lt;Character&gt; list = Arrays.asList('a', 'b', 'c');
String s = list.toString().replaceAll("[,\\s\\[\\]]", "");</code>

blocks|key|869776|text|+private+void+countChar()+throws+IOException+{
++++HashMap+hashMap+=+new+HashMap();
++++List+list+=+new+ArrayList();
++++list+=+"aammit".chars().mapToObj(r+->+(char)+r).collect(Collectors.toList());
++++list.stream().forEach(e+->+{
++++++++hashMap.computeIfPresent(e,+(K,+V)+->+(int)+V+%2B+1);
++++++++hashMap.computeIfAbsent(e,+(V)+->+1);
++++});

++++System.out.println(hashMap);

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|869777|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> private void countChar() throws IOException {
 HashMap hashMap = new HashMap();
 List list = new ArrayList();
 list = "aammit".chars().mapToObj(r -&gt; (char) r).collect(Collectors.toList());
 list.stream().forEach(e -&gt; {
 hashMap.computeIfPresent(e, (K, V) -&gt; (int) V + 1);
 hashMap.computeIfAbsent(e, (V) -&gt; 1);
 });

 System.out.println(hashMap);

}
</code></pre>

blocks|key|652532|text|使用Joiner类的join：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|652533|//+create+character+list+and+initialize+
List<Character>+arr+=+Arrays.asList('a',+'b',+'c');+++
String+str+=+Joiner.on("").join(arr);
System.out.println(str);|code-block|syntax|javascript|652534|使用toString，然后删除,+and+spaces|652535|import+com.google.common.base.Joiner;+

....
<Character>+arr+=+Arrays.asList('h',+'e',+'l',+'l',+'o');+
//+remove+[]+and+spaces+
String+str+=+arr.toString()+
++++++++++.substring(1,+3+*+str.size()+-+1)+//3+bcs+of+commas+,
++++++++++.replaceAll(",+",+"");+
System.out.println(str);|652536|或者使用streams：|652537|import+java.util.stream.Collectors;+
...
//+using+collect+and+joining()+method+
String+str+=++arr.stream().map(String::valueOf).collect(Collectors.joining());+|652538|entityMap^0|2|6|A|4|0|0|2|8|F|C|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|16|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

Using <code>join</code> of a <code>Joiner</code> class:

<pre><code>// create character list and initialize 
List&lt;Character&gt; arr = Arrays.asList('a', 'b', 'c'); 
String str = Joiner.on("").join(arr);
System.out.println(str);
</code></pre>

Use <code>toString</code> then remove <code>, and spaces</code>

<pre><code>import com.google.common.base.Joiner; 

....
&lt;Character&gt; arr = Arrays.asList('h', 'e', 'l', 'l', 'o'); 
// remove [] and spaces 
String str = arr.toString() 
 .substring(1, 3 * str.size() - 1) //3 bcs of commas ,
 .replaceAll(", ", ""); 
System.out.println(str);
</code></pre>

Or by using streams:

<pre><code>import java.util.stream.Collectors; 
...
// using collect and joining() method 
String str = arr.stream().map(String::valueOf).collect(Collectors.joining()); 
</code></pre>

How to convert an <code>ArrayList&lt;Character&gt;</code> to a <code>String</code> in Java?
The <code>List.toString</code> method returns it as <code>[a,b,c]</code> string - I want to get rid of the brackets (etcetera) and store it as <code>abc</code>.

Converting ArrayList of Characters to a String?

Java

如何在Java中将ArrayList<Character>转换为String？List.toString方法将其作为[a,b,c]字符串返回-我希望去掉方括号(etcetera)并将其存储为abc。

问将字符的ArrayList转换为字符串？
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