Best way to concatenate List of String objects? [duplicate]

Answer 1

Next variation on Peter Lawrey's answer without initialization of a new string every loop turn

String concatList(List<String> sList, String separator)
{
    Iterator<String> iter = sList.iterator();
    StringBuilder sb = new StringBuilder();

    while (iter.hasNext())
    {
        sb.append(iter.next()).append( iter.hasNext() ? separator : "");
    }
    return sb.toString();
}

Answer 2

if you have json in your dependencies.you can use new JSONArray(list).toString()

Answer 3

I prefer String.join(list) in Java 8

Answer 4

ArrayList inherits its toString()-method from AbstractCollection, ie:

public String toString() {
    Iterator<E> i = iterator();
    if (! i.hasNext())
        return "[]";

    StringBuilder sb = new StringBuilder();
    sb.append('[');
    for (;;) {
        E e = i.next();
        sb.append(e == this ? "(this Collection)" : e);
        if (! i.hasNext())
            return sb.append(']').toString();
        sb.append(", ");
    }
}

Building the string yourself will be far more efficient.

---

If you really want to aggregate the strings beforehand in some sort of List, you should provide your own method to efficiently join them, e.g. like this:

static String join(Collection<?> items, String sep) {
    if(items.size() == 0)
        return "";

    String[] strings = new String[items.size()];
    int length = sep.length() * (items.size() - 1);

    int idx = 0;
    for(Object item : items) {
        String str = item.toString();
        strings[idx++] = str;
        length += str.length();
    }

    char[] chars = new char[length];
    int pos = 0;

    for(String str : strings) {
        str.getChars(0, str.length(), chars, pos);
        pos += str.length();

        if(pos < length) {
            sep.getChars(0, sep.length(), chars, pos);
            pos += sep.length();
        }
    }

    return new String(chars);
}

Answer 5

Have you seen this Coding Horror blog entry?

The Sad Tragedy of Micro-Optimization Theater

I am not shure whether or not it is "neater", but from a performance-standpoint it probably won't matter much.

Answer 6

Using Java 8+

String str = list.stream().collect(Collectors.joining())

or even

String str = String.join("", list);