blocks|key|3977230|text|希望这个帮助:+Java+8+Stream+&+Collector：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3977231|++++String+word+=+"AAABBB";
++++Map<Character,+Integer>+charCount+=+word.chars().boxed().collect(Collectors.toMap(
++++++++++++++++++++k+->+Character.valueOf((char)+k.intValue()),
++++++++++++++++++++v+->+1,
++++++++++++++++++++Integer::sum));
++++System.out.println(charCount);

Output:
++++{A=3,+B=3}|code-block|syntax|javascript|3977232|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|$]]

Hope this help : Java 8 Stream &amp; Collector:

<pre><code> String word = "AAABBB";
 Map&lt;Character, Integer&gt; charCount = word.chars().boxed().collect(Collectors.toMap(
 k -&gt; Character.valueOf((char) k.intValue()),
 v -&gt; 1,
 Integer::sum));
 System.out.println(charCount);

Output:
 {A=3, B=3}
</code></pre>

blocks|key|3979689|text|下面是另一种简单的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3979690|Map<String,+Integer>+charCount+=+new+HashMap();
++++for(String+charr:+s.split("")){
++++++++charCount.put(charr,charCount.getOrDefault(charr,0)%2B1);
}|code-block|syntax|javascript|3979691|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|$]]

Figured out, below is another simple way.

<pre><code>Map&lt;String, Integer&gt; charCount = new HashMap();
 for(String charr: s.split("")){
 charCount.put(charr,charCount.getOrDefault(charr,0)+1);
}
</code></pre>

blocks|key|3979693|text|+++++String+str+=+"Hello+Manash";
++++Map<Character,Long>+hm+=+str.chars().mapToObj(c->+
++++(char)c).collect(Collectors.groupingBy(c->c,Collectors.counting()));
++++System.out.println(hm);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3979694|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> String str = "Hello Manash";
 Map&lt;Character,Long&gt; hm = str.chars().mapToObj(c-&gt; 
 (char)c).collect(Collectors.groupingBy(c-&gt;c,Collectors.counting()));
 System.out.println(hm);
</code></pre>

blocks|key|3977297|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3977298|++++List<Character>+chars=Arrays.asList('h','e','l','l','o','w','o','r','l','d');
++++Map<Character,Long>+map=chars.stream().map(c->c).
++++collect(Collectors.groupingBy(c->c,Collectors.counting()));
++++System.out.println(map);|code-block|syntax|javascript|3977299|输出：|3977300|{r=1,+d=1,+e=1,+w=1,+h=1,+l=3,+o=2}|3977301|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|$]]

Try this one :
<pre><code> List&lt;Character&gt; chars=Arrays.asList('h','e','l','l','o','w','o','r','l','d');
 Map&lt;Character,Long&gt; map=chars.stream().map(c-&gt;c).
 collect(Collectors.groupingBy(c-&gt;c,Collectors.counting()));
 System.out.println(map);
</code></pre>
output:
<pre><code>{r=1, d=1, e=1, w=1, h=1, l=3, o=2}
</code></pre>

blocks|key|3979743|text|word.chars().mapToObj(c->+(char)c).collect(Collectors.groupingBy(Function.identity(),LinkedHashMap::new,+Collectors.counting()));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3979744|这将为您提供字符出现顺序的字符计数。|unstyled|3979745|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>word.chars().mapToObj(c-&gt; (char)c).collect(Collectors.groupingBy(Function.identity(),LinkedHashMap::new, Collectors.counting()));
</code></pre>
This will give you character count in order of appearance of the character.

blocks|key|3979769|text|尝试以下方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3979770|方法1：|3979771|++++String+str+=+"abcaadcbcb";
++++
++++Map<Character,+Integer>+charCount+=+str.chars()
++++++++++++.boxed()
++++++++++++.collect(toMap(
++++++++++++++++++++k+->+(char)+k.intValue(),
++++++++++++++++++++v+->+1,+++++++++//+1+occurence
++++++++++++++++++++Integer::sum));
++++System.out.println("Char+Counts:\n"+%2B+charCount);|code-block|syntax|javascript|3979772|方法2：|3979773|++++String+str+=+"abcaadcbcb";
++++Map<Character,+Integer>+charCount+=+new+HashMap<>();
++++for+(char+c+:+str.toCharArray())+{
++++++++charCount.merge(c,++++++++++//+key+=+char
++++++++++++++++1,++++++++++++++++++//+value+to+merge
++++++++++++++++Integer::sum);++++++//+counting
++++}
++++System.out.println("Char+Counts:\n"+%2B+charCount);|3979774|输出：|3979775|++++Char+Counts:
++++{a=3,+b=3,+c=3,+d=1}|3979776|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Try the below approaches:
Approach 1:
<pre><code> String str = &quot;abcaadcbcb&quot;;
 
 Map&lt;Character, Integer&gt; charCount = str.chars()
 .boxed()
 .collect(toMap(
 k -&gt; (char) k.intValue(),
 v -&gt; 1, // 1 occurence
 Integer::sum));
 System.out.println(&quot;Char Counts:\n&quot; + charCount);
</code></pre>
Approach 2:
<pre><code> String str = &quot;abcaadcbcb&quot;;
 Map&lt;Character, Integer&gt; charCount = new HashMap&lt;&gt;();
 for (char c : str.toCharArray()) {
 charCount.merge(c, // key = char
 1, // value to merge
 Integer::sum); // counting
 }
 System.out.println(&quot;Char Counts:\n&quot; + charCount);
</code></pre>
Output:
<pre><code> Char Counts:
 {a=3, b=3, c=3, d=1}
</code></pre>

Wondering is there more simple way than computing the character count of a given string as below?

<pre><code>String word = "AAABBB";
 Map&lt;String, Integer&gt; charCount = new HashMap();
 for(String charr: word.split("")){
 Integer added = charCount.putIfAbsent(charr, 1);
 if(added != null)
 charCount.computeIfPresent(charr,(k,v) -&gt; v+1);
 }

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

Java8: Create HashMap with character count of a String

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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想知道是否有比计算给定字符串的字符数更简单的方法，如下所示？String word = "AAABBB"; Map<String, Integer> charCount = new HashMap(); for(String charr: word.split("")){ Integer ad...

问Java8:使用字符串的字符数创建HashMap
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问Java8:使用字符串的字符数创建HashMapEN