将 ArrayList 个字符转换为字符串？

Question

How to convert an ArrayList<Character> to a String in Java?

The List.toString method returns it as [a,b,c] string - I want to get rid of the brackets (etcetera) and store it as abc .

Answer 1

You can iterate through the list and create the string.

String getStringRepresentation(ArrayList<Character> list)
{    
    StringBuilder builder = new StringBuilder(list.size());
    for(Character ch: list)
    {
        builder.append(ch);
    }
    return builder.toString();
}

Setting the capacity of the StringBuilder to the list size is an important optimization. If you don't do this, some of the append calls may trigger an internal resize of the builder.

As an aside, toString() 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.

Answer 2

Here a possible one-line solution using Java8 streams.

a) List of Character objects to String:

String str = chars.stream()
                  .map(e->e.toString())
                  .reduce((acc, e) -> acc  + e)
                  .get();

b) array of chars (char[] chars)

String str = Stream.of(chars)
                   .map(e->new String(e))
                   .reduce((acc, e) -> acc  + e)
                   .get();

UPDATE (following comment below):

a) List of Character objects to String:

String str = chars.stream()
                  .map(e->e.toString())
                  .collect(Collectors.joining());

b) array of chars (char[] chars)

String str = Stream.of(chars)
                   .map(e->new String(e))
                   .collect(Collectors.joining());

---

Note that the map(e->e.toString()) step in the above solutions will create a temporary string for each character in the list. The strings immediately become garbage. So, if the performance of the conversion is a relevant concern, you should consider using the StringBuilder approach instead.

Answer 3

How about this, Building the list

List<Character> charsList = new ArrayList<Character>();
charsList.add('h');
charsList.add('e');
charsList.add('l');
charsList.add('l');
charsList.add('o');

Actual code to get String from List of Character:

String word= new String();
for(char c:charsList){
word= word+ c; 
}
System.out.println(word);

Still learning if there is a misake point out.

Answer 4

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

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

Answer 5

Using join of a Joiner class:

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

Use toString then remove , and spaces

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);

Or by using streams:

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

Answer 6

Assuming you have a following list:

final ArrayList<Character> charsList = new ArrayList<Character>();
charsList.add('h');
charsList.add('e');
charsList.add('l');
charsList.add('l');
charsList.add('o');

This will yield hello (I am using org.apache.commons.lang.ArrayUtils helper class):

final Character[] charactersArray =
    charsList.toArray(new Character[charsList.size()]);
final char[] charsArray = ArrayUtils.toPrimitive(charactersArray);
System.out.println(String.valueOf(charsArray));

Answer 7

a tiny complement to @waggledans 's answer

a) List of Character objects to String:

String str = chars.stream().map(e->e.toString()).collect(Collectors.joining());

which e->e.toString() can be replaced by Object::toString

String str = chars.stream().map(Object::toString).collect(Collectors.joining());

Answer 8

Easiest is to loop through.

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);

Answer 9

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.

Answer 10

I consider this an easy and smart way

 // given list containing the chars
 List<Character> arr = Arrays.asList('a', 'b', 'c'); 

// convert to string.
String str = arr.toString(); // result is "[abc]"

// get rid of the start and the end char i.e '[' & ']'
arr = arr.substring(1, tmpArr.length()-1);  //result is "abc"

Answer 11

I would say:

public String arayListToString(ArrayList arrayList){

StringBuffer b = new StringBuffer();

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

return b.toString();
}

Answer 12

 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 + 1);
        hashMap.computeIfAbsent(e, (V) -> 1);
    });

    System.out.println(hashMap);

}